A recent series of questions from one student involved interesting combinations of trigonometric identities and solutions of polynomials. At one time using trig to solve equations was far better known than it is today, and these presumably are meant in part as an introduction to those ideas. Some were quite challenging to us as we tried to solve unfamiliar problems.

The questions came from Giridharan last October and January. Here is the first:

If cos4θ = ½, prove that the roots of the equation 16c

^{4 }– 16c² + 1 = 0 are cos(π/12), cos(5π/12), cos(7π/12), and cos(11π/12). Hence prove that the roots of the equation 16x² – 16x + 1 = 0 are cos²(π/12) and cos²(5π/12).My working:

Given cos4θ = ½

⇒ 8 cos

^{4}θ – 8 cos²θ + 1 = ½⇒ 16 cos

^{4}θ – 16 cos²θ + 1 = 0⇒ 16 c

^{4 }– 16 c² + 1 = 0 where c = cosθI don’t know how to proceed further. Kindly help me

Here, unlike the usual problem, we are not asked to solve an equation, only to show how some very strong hints lead to a solution. Presumably the equation was designed with this solution in mind, and the purpose of the problem is to stretch the students’ minds by exploring a very specialized and different way to approach polynomial equations.

There are two parts. First, we are to show that certain functions of a certain angle are zeros of a certain 4th-degree polynomial. Second, we are to use that fact to solve a mere *quadratic* equation – that is, to show that the roots, which we could have found much more easily as radicals, are in fact specific trig functions.

Giridharan started with a known formula for 4 times an angle, which can for example be found in the *Ask Dr. Math* FAQ:

cos(4x) = 8 cos^{4}(x) - 8 cos^{2}(x) + 1

(You can prove this by applying a double-angle formula twice.)

I answered, seeing that he was almost there:

Hi, Giridharan.

You’ve done the hard part.

Now just solve cos4θ = ½ the usual way, first finding 4θ. You should find that the solution looks like the goal of the problem you were given.

It was odd that *θ* never appeared in the problem after the first mention; that is because the use of cos(4*θ*) was merely a hint to the rest, not a definition of a variable that was part of the problem itself. It was never meant to last long as a variable! So finding what *θ* actually is has to be part of the work.

He did that:

Thank you Sir

Is my working correct after this?

Given cos4θ = ½

⇒ 4θ = cos

^{-1}(½)⇒ θ = ¼ cos

^{-1}(½)= ¼[(π/3) + 2nπ, (5π/3) + 2nπ]

when n = 0, θ = (π/12), (5π/12)

when n = 1, θ = (7π/12), (11π/12)

After this θ value repeats.

Hence roots of the given equation are cos(π/12), cos(5π/12), cos(7π/12), and cos(11π/12).

How to solve roots of the equation 16x² – 16x + 1 = 0 are cos²(π/12) and cos²(5π/12)?

He has used “\(\cos^{-1}\) to mean not just the *principal value* of the inverse cosine, but *any* angle with the indicated cosine; so the possible values of *θ* are \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}\), and coterminal angles. This completes the first part of the problem. Nothing was really difficult except determining what this unusual question wanted.

I responded, again giving broad hints because his work suggested that was all that was needed:

Very good.

Technically, it is not right to say that θ repeats after the four values, but its cosine does, which is what we are interested in.

Now go back to your earlier work, where you showed that 16c

^{4 }– 16c^{2 }+ 1 = 0 where c = cosθ (for any of the 4 values you just found for θ). That is, cos(π/12), cos(5π/12), cos(7π/12), and cos(11π/12) are the four roots of 16c^{4 }– 16c^{2 }+ 1 = 0.Another way to try to solve 16c

^{4 }– 16c^{2 }+ 1 = 0 is by substitution, letting u = c^{2}. What can you do with that idea?

He answered, just making the substitution I suggested, but not following through:

Let x = c²

Then the original equation becomes 16x² – 16x + 1 = 0.

Sir am I missing some results from theory of equations?

He probably expects something more complicated that what he has to do. The quartic and the quadratic are equivalent, and knowing the roots of one tells us the roots of the other. Normally we would be using the quadratic to find the roots of the quartic in radical form; what we are told to do here is to reverse that, using the roots of the quartic equation to find the roots of the quadratic in trig form!

I explained that:

How would you try solving 16c

^{4 }– 16c^{2 }+ 1 = 0 for c, using substitution? You would solve 16x² – 16x + 1 = 0 for x, and then solve x = c² for c, for each of those values of x.So, for example, if c = cos(π/12) is a solution of 16c

^{4 }– 16c^{2 }+ 1 = 0, what does that tell you about a solution of 16x² – 16x + 1 = 0?

He did so, and just stated the results:

Then cos²(π/12) and cos²(5π/12) are the roots of the equation 16x² – 16x + 1 = 0.

Am I correct Sir?

Let’s do that ourselves. Since the roots of \(16c^4 – 16c^2 + 1 = 0\) are (from the first part of the problem) $$c = \cos(\frac{\pi}{12}), \cos(\frac{5\pi}{12}), \cos(\frac{7\pi}{12}), \text{ and }\cos(\frac{11\pi}{12})$$, the substitution \(x = c^2\) implies that the solutions in *x* are $$x = \cos^2(\frac{\pi}{12}), \cos^2(\frac{5\pi}{12}), \cos^2(\frac{7\pi}{12}), \text{ and }\cos^2(\frac{11\pi}{12}).$$

But … these are *four* solutions, not the *two* solutions we expect from a quadratic equation! What do we do about that?

It turns out that the last two of them are duplicates. For example, $$\cos(\frac{7\pi}{12}) = \cos(\pi – \frac{5\pi}{12}) = -\cos(\frac{5\pi}{12}),$$ so $$\cos^2(\frac{7\pi}{12}) = \cos^2(\frac{5\pi}{12}).$$ So the two solutions are in fact $$x = \cos^2(\frac{\pi}{12})\text{ and }\cos^2(\frac{5\pi}{12})$$.

I checked that Giridharan had not just glossed over this issue, and we were done.

A couple months later, we got a similar, but more difficult, question from Giridharan:

Express sin9θ/sinθ as a polynomial in cosθ and deduce that

(i) sec²(π/9) + sec²(2π/9) + sec²(4π/9) = 36

I have expressed sin9θ/sinθ

=256cos

^{8}θ – 448cos^{6}θ + 240cos^{4}θ – 49cos²θ + 1After this I am not able to proceed further. Kindly help me.

Before proceeding, let’s carry out this first part. I can’t be sure what Giridharan started with, but one way to do this is to know (or derive, or look up) these formulas for a triple angle (for example, from the Ask Dr. Math FAQ):

sin(3x) = 3 sin(x) - 4 sin^{3}(x) cos(3x) = 4 cos^{3}(x) - 3 cos(x)

Applying it twice, $$\sin(9x) = \sin(3(3x)) = 3\sin(3x) – 4\sin^3(3x) = 3(3\sin(x) – 4\sin^3(x)) – 4(3\sin(x) – 4\sin^3(x))^3$$

Expanding the cube of a binomial using the formula $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ we get $$\sin(9x) = 9\sin(x) – 12\sin^3(x) – 4(27\sin^3(x) – 36\sin^5(x) + 48\sin^7(x) – 64\sin^9(x))\\ = 9\sin(x) – 120\sin^3(x) + 144\sin^5(x) – 208\sin^7(x) + 256\sin^9(x)$$

But we were told to express this in terms of the cosine, not the sine! (Oops.) Let’s back up and express our triple angle formula in terms of cosine (as far as we can):

sin(3x) = sin(x)(3 - 4 sin^{2}(x)) = sin(x)(3 - 4 (1 - cos^{2}(x))) = sin(x)(4 cos^{2}(x) - 1)

Now let’s apply this formula twice: $$\sin(9x) = \sin(3(3x)) = \sin(3x)(4\cos^2(3x) – 1) = \sin(x)(4\cos^2(x) – 1)(4(4\cos^3(x) – 3\cos(x))^2 – 1)\\ = \sin(x)(4\cos^2(x) – 1)(4(16\cos^6(x) – 24\cos^4(x) + 9\cos^2(x)) – 1))\\ = \sin(x)(4\cos^2(x) – 1)(64\cos^6(x) – 96\cos^4(x) + 36\cos^2(x) – 1)\\ = \sin(x)(256\cos^8(x) – 384\cos^6(x) + 144\cos^4(x) – 4\cos^2(x) – 64\cos^6(x) + 96\cos^4(x) – 36\cos^2(x) + 1)\\ = \sin(x)(256\cos^8(x) – 448\cos^6(x) + 240\cos^4(x) – 40\cos^2(x) + 1)$$

Divide by the sine, and we have almost what Giridharan got: $$\frac{\sin(9x)}{\sin(x)} = 256\cos^8(x) – 448\cos^6(x) + 240\cos^4(x) – 40\cos^2(x) + 1$$

So I answered:

Hi, Giridharan.

Thanks for showing the work you’ve done. It’s a good start.

There is one small error in your work; it should be:

sin9θ/sinθ = 256cos

^{8}θ – 448cos^{6}θ + 240cos^{4}θ –40cos^{2}θ + 1I’ve confirmed this by graphing both sides.

Here is that graph (done on Desmos); red is the LHS and blue dots are the RHS):

When I tell someone they are wrong, I like to be sure!

I have not finished the problem, but I would expect that in some way you have to apply this fact to

θ = π/9,

θ = 2π/9, and

θ = 4π/9,

since in each of these cases sin 9θ = 0.

One way to apply this would be to rearrange the equation

256cos

^{8}θ – 448cos^{6}θ + 240cos^{4}θ – 40cos^{2}θ + 1 = 0to

1 = -256cos

^{8}θ + 448cos^{6}θ – 240cos^{4}θ + 40cos^{2}θ1/cos

^{2}θ = -256cos^{6}θ + 448cos^{4}θ – 240cos^{2}θ + 40This is true for each of the three values I listed,

See what you can do with this start.

It seemed clear that the identity should be applied to the three numbers used in the second part of the problem, but my suggestion beyond that was at this point a wild guess, based on the need to transform cosines to secants. Sometimes when you don’t know what to do next, it is not unreasonable to aim for something specific “on the horizon”, if only to get you moving somewhere; you can adjust later. And this did need an adjustment!

I wrote that late at night (which accounts for my not trying harder to take it further). In the middle of the next day (my time), Giridharan responded,

Dear Sir

Thank you for correcting my mistake. I am working on your hint

If my hint was bad, I needed to correct it. While waiting for a further response (until late the next day), I worked enough to have a definite answer. At that point I wrote back:

I went back to look at the problem again, and solved it. The rearrangement I suggested last time is not useful.

I first checked with a calculator to make sure the claim is correct, in case you had copied something wrong; it is correct.

Then I considered this:

256cos

^{8}θ – 448cos^{6}θ + 240cos^{4}θ – 40cos^{2}θ + 1 = 0means that cosθ is a root of the equation

256x

^{8}– 448x^{6}+ 240x^{4}– 40x^{2}+ 1 = 0.But that led to further thoughts. There are things we know about roots of a polynomial; but this one has not 3 but 8 roots.

What are the others, in addition to the three that are mentioned in the problem? Can we use the fact thatall exponents are even? How are thethree fractionsin the statement to be proved related to the polynomial?Thoughts like these led to an answer; I am leaving some important ideas for you to have the fun of discovering.

Why, I wondered, would a question about the roots of an eighth-degree polynomial only involve three numbers, not all eight roots. I might also have thought back to the first problem we looked at above, where only half the roots were needed …

Keep in mind that the problem is to show that $$\sec^2\left(\frac{\pi}{9}\right) + \sec^2\left(\frac{2\pi}{9}\right) + \sec^2\left(\frac{4\pi}{9}\right) = 36,$$ that is, $$\frac{1}{\cos^2\left(\frac{\pi}{9}\right)} + \frac{1}{\cos^2\left(\frac{2\pi}{9}\right)} + \frac{1}{\cos^2\left(\frac{4\pi}{9}\right)} = 36$$

I hoped this was a strong enough hint!

He replied, having done some very good thinking:

Sir

Kindly let me know if my following steps are correct:

Let θ=π/9, 2π/9, 3π/9, and 4π/9; in each of these cases sin9θ = 0.

256x

^{8}– 448x^{6}+ 240x^{4}– 40x^{2}+ 1 = 0 where x = cosθLet x² = 1/z; then the above equation reduces to

z

^{4 }– 40 z³ + 240 z² – 448 z + 256 = 0hence

sec²(π/9) + sec²(2π/9) + sec²(3π/9) + sec²(4π/9) = 40

⇒ sec²(π/9) + sec²(2π/9) + 4 + sec²(4π/9) = 40

⇒ sec²(π/9) + sec²(2π/9) + sec²(4π/9) = 36

Prod of the roots:

sec²(π/9) sec²(2π/9) sec²(3π/9) sec²(4π/9) = 256

sec²(π/9) sec²(2π/9) 4 sec²(4π/9) = 256

⇒ sec(π/9) sec(2π/9) sec(4π/9) = 8

So he first expanded his view from 3 numbers to 4 (the positive roots of the polynomial), by including π/3 (whose cosine is simple), then made a change of variables to reduce the degree to 4, and then used the fact that the sum of the zeros of a monic polynomial is the negative of the coefficient of the term after the leading term. Or, more generally, the sum of the roots of \(ax^n + bx^{n-1} + … + z = 0\) is \(\frac{-b}{a}\).

I suspect the last part he showed, about the products, may have been part (ii) of the problem.

I replied, explaining my slightly more complicated approach, without the substitution:

Excellent.

What I did was just a little different.

When I realized that

the other rootis sec²(3π/9) = 4, sothe sum of that and the other numbers is 40, that encouraged me to pursue the roots of the equation.But rather than let x^2 = 1/z as you (wisely!) did, I took a smaller step. I saw merely that cos²(π/9), cos²(2π/9), cos²(3π/9), and cos²(4π/9) are the roots of 256x

^{4}– 448x^{3}+ 240x^{2}– 40x + 1 = 0, and asked myself whether I could show that.the sum of their reciprocalsis 40If the roots are a, b, c, d, then 1/a+1/b+1/c+1/d = (bcd+acd+abd+abc)/(abcd) = 40/1 = 40, because the coefficient of the linear term is the negative of the sum of products of all but one root.

I like your way more. But then, I had to make up my own hint …

Good work.

So rather than use the fact about the sum of roots and the second-from-left coefficient, I used a similar fact about the second-from-right coefficient, which was the long way around.

Giridharan closed:

Thank you Sir for your encouraging words. Every time I correspond with you I learn something different.

Continue to guide me in future also.

The next day Giridharan completed the trilogy with this very similar question:

Expand cos6θ as a polynomial in cosθ.

Deduce that the roots of the equation 64x³ – 96x² + 36x – 3 = 0 are cos²(π/18), cos²(5π/18), and cos²(7π/18).

I expanded cos6θ = 32 cos

^{6}θ – 48 cos^{4}θ + 18 cos²θ – 1.Let cos6θ = ½;

then θ = π/18, 5π/18, 7π/18, 11π/18, 13π/18, and 17π/18.

Sir is my approach correct?

I replied,

Yes, that is correct.

You have just a little more work to show to complete the proof.

He came back after a few days:

cos6θ = ½

⇒ 2 cos6θ = 1

⇒ 2 [32 cos6θ – 48 cos4θ + 18 cos²θ – 1] = 1

⇒ 64 cos

^{6}θ – 96 cos^{4}θ + 36 cos²θ – 2 = 1⇒ 64 cos

^{6}θ – 96 cos^{4}θ + 36 cos²θ – 3 = 0 ––– (1)-cos(11π/18) = cos(π – 11π/18) = cos(7π/18)

-cos(13π/18) = cos(π – 13π/18) = cos(5π/18)

-cos(17π/18) = cos(π – 17π/18) = cos(π/18)

Put cos²θ = x in (1)

Then (1) becomes 64x³ – 96x² + 36x – 3 = 0

Hence cos²(π/18), cos²(5π/18), and cos²(7π/18) are the roots of the above equation.

Sir is it ok?

Yes, it was; I responded only by adding the sort of detail I often add in my commentaries here in the blog so that other readers can follow, because he had it all right:

Yes, assuming you have previously demonstrated the expansion of cos6θ.

I like putting a few more words into my proofs, as if I were teaching rather than communicating to an experienced mathematician!

With more words, we might say that, knowing that

cos6θ = 32 cos

^{6}θ – 48 cos^{4}θ + 18 cos²θ – 1,the equation cos6θ = ½ is equivalent to

64 cos

^{6}θ – 96 cos^{4}θ + 36 cos²θ – 3 = 0,which in turn is equivalent to

64x³ – 96x² + 36x – 3 = 0 if x = cos²θ;

since the solution set of cos6θ = ½ is

{π/18, 5π/18, 7π/18, 11π/18, 13π/18, 17π/18},

and cos²(11π/18), cos²(13π/18), and cos²(17π/18) are the same numbers as cos²(π/18), cos²(5π/18), and cos²(7π/18), therefore the solution set of 64x³ – 96x² + 36x – 3 = 0 is

{cos²(π/18), cos²(5π/18), cos²(7π/18)}.

But that is what you are saying.

Another interesting problem!

]]>Let’s start with this real application from 1999:

Will the Tree Hit the House? There is a tree out in front of our yard. It is tilted slightly at 70 degrees. Our house is 66 1/2 feet away from the tree. The angle from our house to the top of the tree is 40 degrees. My family is worried that if we have a big storm the tree will fall and hit the house. I read somewhere that if you have 2 angles and a side you can figure out the dimensions of the triangle. I haven't taken trig yet, so could you please help me out? Thanks. Billy

Presumably the tree is tilted toward the house, so it would fall in that direction:

We’ll assume, as in the picture, that the angle to the top of tree was measured from the ground; if it was measured at eye level, say 5 feet up, then the tree would be about 5 feet taller than our calculation:

In word problems, we generally don’t worry about real-life details like this, but if this is a real decision that has to be made, then we should.

Doctor Rick answered, with a quick solution, and ignoring my little issue:

Hi, Billy. Nice question.You don't need trigonometryto answer your basic question, which is,could the tree hit the house? There is a theorem in geometry (it's Euclid's Proposition 19) that says: "In any triangle the side opposite the greater angle is greater." Let's see how we can use this. Here is a figure: B /\ / \ TREE / \ / \ / \ / 70 40 \ /________________________\ HOUSE A 66.5' C I am assuming that the tree is tilted toward the house. Now, what is the angle at the top of the triangle, angle B? Since the sum of the angles in a triangle is 180 degrees, that angle is 70 degrees.

Here is the picture, abstracted from the problem itself, labeled, and with the third angle labeled, ready to work on:

Now we can use the theorem. The distance to the house, AC, is opposite a 70-degree angle. The height (or rather length) of the tree, AB, is opposite a 40-degree angle. The side opposite the greater angle is greater, so AC is greater than the length of the tree AB. The tree cannot hit the house.

It’s a good idea to look for shortcuts like this that save work. This is also one reason we ask students to tell us the context of their problem, not just the bit they are working on: If Billy had just given us the triangle and asked for the length of side *c*, we would not have noticed that he didn’t really need that.

But, of course, it will be interesting to actually find the height of the tree, and that will introduce Billy to some new ideas, so why not!

You are correct that trigonometry can be used to find the actual length of the tree. We use the Law of Sines, which puts numbers into Euclid's theorem. AB/sin(C) = AC/sin(B) AB = AC * sin(C)/sin(B) = 66.5' * sin(40)/sin(70) The sine of 40 degrees, abbreviated sin(40), is 0.642787610 from my calculator. The sine of 70 degrees is 0.939692621. Therefore AB = 66.5' * 0.642787610 / 0.939692621 = 45.5' So you have 21 feet to spare.

The comment, “which puts numbers into Euclid’s theorem” is of interest. If you don’t follow that, the idea is that the Law of Sines, in saying that sides are proportional to sines of angles (within a given triangle), tells us in particular that the longest side is opposite the angle with the largest *sine*. But since the sine is an increasing function (on acute angles), this amounts to a more precise version of Euclid’s theorem that the longest side is opposite the largest *angle*.

(What about obtuse triangles, you ask? In that case, the longest side is opposite the obtuse angle, and although the sine is a decreasing function for obtuse angles, its sine is still greater than either other angle, because the supplement of the obtuse angle is equal to the sum of the acute angles. But we digress …)

Notice that the 5 feet or so that would be added if the measurement was made at eye level doesn’t affect the conclusion, that the house is safe. This couldn’t have been determined by the inequality method.

But did you notice another special fact about the problem that allows us to solve the problem in a different way?

Since the triangle turns out to be isosceles, we can split it into two congruent right triangles:

Half the height of the tree is *x*, which we can calculate as \(x = 66.5\cos(70°) = 22.74\). Doubling this gives the height as 45.5 feet, the same as before.

Here is a more typical classroom problem, from 1995 (a couple months after *Ask Dr. Math* started):

How Tall is the Building on the Hill? A building is 30 meters high and is on top of a hill. The angles of elevation of the top and bottom of the building from a point at the foot of the hill are 64 degrees and 58 degrees 4 feet respectively. How high is the hill?

Picture it like this (the angles are not exact):

Doctor Margaret answered:

Hi Meredith!! Thanks for writing Dr. Math. I am not sure exactly what your question means. Maybe you could clarify: What do you mean by "58 degrees 4 feet"? Is there more to the problem?

I suspect the problem may have said 58° 4′, meaning 58 degrees, 4 minutes (though that seems like an oddly precise angle in context). We’ll ignore the 4.

If I ignore that part of the problem, this what we get: | 30m | we can draw two triangles - one connecting | the point at the bottom to the top of the | building and one to the bottom. | --- \ |\ \ | \ hill \ | \ \ |\ | \ \ | \ 30+x | \ x | \ | \ | \ | \ |____\58 |_______\64 y y The height of the hill is x, and the measures of the angles at the base of the triangles are 58 and 64. Since these are right triangles, we know that the measures of the angles at the tops are their complements.

Here is a clearer picture keeping all the parts together, and labeling them:

Doctor Margaret’s approach of drawing each triangle separately is an excellent one, after first seeing how all the parts are related in context. It allows us to focus on each triangle alone, abstractly:

Then we can use the Law of Sines to form two equations and two unknowns. Do you know about the Law of Sines? In any triangle with sides of lengths A, B, and C, and the angles opposite those sides, a, b, and c. Then a b c ----- = ----- = ----- sin A sin B sin C For the first triangle, it is true that 30+x y ----- = ----- sin 64 sin 26 Do the same thing for the second triangle and then solve for x.

Now, Doctor Margaret is applying the Law of Sines to right triangles, where it is really unnecessary. We could have just used (in the larger triangle) $$\tan(64°) = \frac{x+30}{y}$$ and (in the smaller triangle) $$\tan(58°) = \frac{x}{y}.$$ These are equivalent to Doctor Margaret’s equations, when you consider that $$\tan(64°) = \frac{\sin(64°)}{\cos(64°)} = \frac{\sin(64°)}{\sin(26°)}.$$

Solving each equation for *y* and setting them equal, we have $$\frac{x+30}{\tan(64°)} = \frac{x}{\tan(58°)}$$ Cross-multiplying, we have $$(x+30)\tan(58°) = x\tan(64°)$$ from which we obtain $$x = \frac{30\tan(58°)}{\tan(64°)-\tan(58°)} = 106.7$$

What if we really use the Law of Sines, applying it to our oblique triangle ABC? If we can find either of the diagonal lines (lines of sight), then we can use one of the right triangles we saw above to find the height.

We can do a little subtraction of angles, and we’ll know all three angles and one side:

Using the Law of Sines, $$\frac{z}{\sin(26°)} = \frac{30}{\sin(6°)},$$ so $$z = \frac{30\sin(26°)}{\sin(6°)} = 125.81$$

Then $$x = z\sin(58°) = 106.7$$ as we saw the other way.

Meredith answered,

Thanks for the help. The answer you gave me fit the problem although my teacher went about a different way of getting it. I will probably write again soon. Thanks!

My guess would be that the teacher used the two-right-triangle method without the Law of Sines, but who knows!

Let’s do one more, from 2001, which amounts to that last one turned on its side and solving for a different part:

Height of the Plane Question: Towers A and B are known to be 4.1 mi. apart on level ground. A pilot measures theangles of depressionto the towers to be 36.5 degrees and 25 degrees, respectively.Find the heightof the airplane. I tried to answer this question and simply did not have any idea how to do it.

Here is the problem:

Recall that an “angle of depression” means the angle by which you need to “depress” your vision (look down from the horizontal) to see something. Also, though it says the angles “to the towers”, that must refer to the *base* of each tower, or we would have to take their height(s) into account.

Doctor Jaffee answered:

Hi Peter, A good picture can be very helpful in solving a problem like this one. Draw line PQ, where P and Q represent the bases of the two towers and the length of the segment PQ is 4.1 miles. Then draw line AB parallel to line PQ. Point A represents the location of the airplane. Point A should be situated so that the measure of angle BAQ is 25, and the measure of angle BAP is 36.5. Finally, locate point C on line PQ so that AC is perpendicular to PQ.

Here is his picture, again abstracted from the problem and labeled:

I added in the fact that the angles at P and Q are congruent to the respective angles at A because they are “alternate interior angles” on lines crossing the two parallel lines AB and PQ. I also subtracted to find angle PAQ, which will be important.

Now you should be able to calculate the measures of all the angles in the drawing fairly easily, and then use the Law of Sines in triangle APQ to calculate the length of AP. Once you know that, you should be able to work in triangle ACP and calculate AC, the height of the airplane.

He only described the procedure for solving this, using the oblique triangle as above. Let’s finish it:

In order to find *h*, we want to find *y* and use right triangle ACP. To find *y*, we use the Law of Sines: $$\frac{y}{\sin(25°)} = \frac{4.1}{\sin(11.5°)}$$ Solving, we have $$y = \frac{4.1\sin(25°)}{\sin(11.5°)} = 8.69$$

Now $$h = y\sin(36.5°) = 8.69\sin(36.5°) = 5.17\text{ mi}$$

A similar problem was asked in 2003 without any actual measurements:

Determine Flagpole Height without Access to the Pole

Doctor Douglas proposed essentially the same method.

I’ll leave it to you, the reader, to try other methods, such as our two-right-triangle approach.

]]>Some kinds of problems can be solved at various levels; in particular, when we get a problem about ratios, we often can’t be sure whether the student knows only arithmetic, or can use algebra, which usually makes the problem easier. This is one reason we ask for a category (such as Arithmetic or Algebra), and also for an age range. It’s even more helpful when we see the start of a solution, which is the biggest hint to the appropriate method to use. This one led to an interesting discussion.

The problem came in last October, from Eric, who put himself in the 10-13 year old category, where students may or may not be doing algebra:

The problem is as follows:

Tom and Jerry paid for a friend’s present in the ratio 7:4. In doing so, Jerry spent 1/4 of his money and Tom had $99 left. If the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present, how much did the present cost?

To be honest, I don’t know where to begin. It throws so much information at you, that is broken and detached from the rest of the problem, and it’s like you need to put it all together and add some pieces. I had a go, then my mom had a go, and we both got nowhere than just writing down the information and staring at the answer key, which didn’t explain how to solve the question. If I could have some intel on

how to start my thinkingfor this andhow everything fits togetherthat would be fantastic.If you want to know the answer it is:

$33

That isn’t a simple problem is it? This will be a lesson on how to gather information and make sense of it.

Answer keys often are not helpful for telling how to solve a problem; but they can help us see whether we are interpreting a problem correctly, and whether the student is confused only because the given answer is wrong. (It happens!) Here, it turns out to be right.

Doctor Rick took the question:

Hi, Eric, thanks for writing to The Math Doctors!

The fact that you put your question in the Algebra category tells me that

you expect to be using variables and solving equations. That’s a big help —if I had to tackle the problem solely in terms of ratios, I don’t know what I’d do!Algebra helps us to encode the information in a complex problem like this into variables and equations; once we’ve done that, I can sit back and just solve the equations, and we’ll be done!

Many problems, including ratios, can be solved in creative ways without algebra; what algebra does is to take out the need for creativity and make the main work routine. Well, mostly …

Here’s a suggestion to get you started. I see three main numbers that we don’t know, but that we want to talk about. So we can

define three variablesto represent these quantities: let’s sayT = the amount of money that Tom had to start with (in dollars)

J = the amount of money that Jerry had to start with (in dollars)

P = the cost of the present (in dollars)

Now

we hope that we can write three equations using these three variables. It sometimes turns out that this isn’t necessary, at least when we’re only interested in finding the value of one of our variables (as here, where we want to solve for P). You will find, though, that youcanwrite three equations.

Note how carefully he has defined the variables. It isn’t just “T = Tom”, as many students write; rather, T is an amount of money that Tom has at a particular time in the problem. Doing this makes it easier to keep track of the details. Orderliness is the key to handling a complex problem.

Now I’d like you to give it a try. Can you come up with

an equation to represent the information in the first sentence? Can you write two more equations from the other information in the problem? Let me see what you can do, even if it isn’t much or if you aren’t confident that you did it right — because that’s a start, and then we’ll have more to talk about.

We like to get the student involved in carrying out the work, so they’ll have the experience to do it again on the next problem, and the next. If we just demonstrated solutions, we’d be doing no more than a textbook does, without the necessary interaction. (That’s also why you may want to submit a question to us rather than read our pre-solved problems.)

Eric responded with a multi-part reply:

Sorry for the

late reply…Thanks for responding so quickly Dr. Rick! I… tried my best with the equations and came up with all sorts of junk like P=T/J, P=7/4, P=T/J-7/4, but then mom was there to save the day, with

7/4=(T-99)/(J/4), which I know isn’t what you asked, but my mom and I were in this together. We spent hours thinking about it. We were invested in this.…

The reply came 50 minutes after Doctor Rick wrote, not exactly late. They did some good work in that time!

Anyway, I came up with the next equation (using your variables thank you so much ahhhhhhhhhhhh) that was

T/J=5/2. Groundbreaking, I know, but it played a good role in the final result!The final equation was

P=J/4+(T-99), created out of nothingness by my mom… again…And so we solved it! From T/J=5/2 we got T=5J/2 and plugged that into the first equation and then the last.

Thank you so much for your help with this problem!

We’ll be seeing how to do this without miraculous intervention.

P.S.: I put this under

Algebrabecause I thought it was the difficulty range for this problem… but it seems that all’s well that ends well…

Is Algebra a difficulty range? It does describe a level of understanding, but it’s largely a way of looking at a problem.

P.P.S.: I know that we already solved this together, but I was wondering:

how would you go about this problem with ratios?This problem was listed in the ratio section of my math workbook. I know that you said that you wouldn’t know where to begin with using ratios, and I am totally fine with you not knowing how to use ratios for this problem! It’s just the thought of using ratios was nagging at the back of my mind for a while… It’s fine though! We got the answer (which was Jerry spending 12 dollars and Tom spending 21 dollars)!

We’ll see that the comment was misunderstood.

Doctor Rick answered each piece, starting with the “lateness”:

Hi, Eric. I don’t consider a response sent 50 minutes after I wrote you to be “late” (but then I’m from the old generation that remembers writing letters) … you seem to have worked quite hard in that time. Also, you probably live not far from my time zone since I received your last message at almost 10:00 pm. I live on the east coast of the USA. Now it’s morning and I’m ready to help

And our generation also remembers submitting a computer program on a stack of cards and waiting a day to get the results …

If you know algebra and you are expected to use it to solve problems, then

“algebra” is the correct category to use. Subject matter matters more than difficulty level.Anyway, let’s talk about the problem and your solution. You (and your mom) did good work. What I said was that I wouldn’t know what to do if I had to solve the problem

in terms of ratios,solelyusing algebra. However, we must certainlywithoutuseratios because two pieces of information arestatedin terms of ratios.

Before we close, I will try solving the problem with only ratios and arithmetic (and a picture), as is taught at the elementary level. But for our purposes, the problem is, at heart, an algebra problem about ratios.

You sound like you were able to come up with one equation, but your mom got the other two, and you aren’t sure how she got them. Let’s go over the problem in detail to see

how you can develop the equations yourself.I defined three variables:

T = the amount of money that Tom had to start with (in dollars)

J = the amount of money that Jerry had to start with (in dollars)

P = the cost of the present (in dollars)

The first sentence, “

Tom and Jerry paid for a friend’s present in the ratio 7:4,” can’t be written yet – I shouldn’t have asked you about that to start with. Before we can write an equation to represent that, we need to find expressions in terms of T, J, and P for these two quantities:the amount of money Tom paid for the present (in dollars)

the amount of money Jerry paid for the present (in dollars)

When you try to write an equation all at once, it is easy to make mistakes. It is best to take it slowly, phrase by phrase. Here we want to write an expression for the ratio of the two amounts paid; since these are not variables, and no information given yet tells us how much was paid, we could only write an equation yet if we defined two more variables for these quantities. Since (having preread the problem before starting to work on it) we know that more information is coming, we just put this on hold for now, and wait patiently.

So we look further into the problem, and find this: “

Jerry spent 1/4 of his money and Tom had $99 left.” If Jerry spent 1/4 of his money on the present, then he spent (1/4)J. If Tom had 99 dollars left, then he spent (T – 99) dollars, since subtracting (T – 99) from T leaves 99. Thus we can now “translate” the two expressions above into algebra:the amount of money Tom paid for the present (in dollars) = T – 99

the amount of money Jerry paid for the present (in dollars) = J / 4

As soon as we write this, we should be checking that it makes sense. So we look at the expression we wrote, *T* – 99; that means $99 less than *T*, which is the amount Ted had; that makes sense. Similarly, the expression *J */ 4 means 1/4 of what Jerry started with, which is just what the problem says he spent. (We need to read carefully and make sure that’s what he spent, not what he had left, which a different problem might have said.)

And now we can translate the first sentence:

Tom and Jerry paid for a friend’s present in the ratio 7:4.

The ratio of what Tom paid to what Jerry paid is 7:4.

(T – 99) : (J / 4) = 7:4

T – 99 = (7 / 4)(J / 4)

The first version is written in terms of ratios, using the colon; then it has been rewritten in more algebraic form, and the division restated as a multiplication (that is, both sides were multiplied by *J* / 4). One reason for doing the latter is that a complex fraction (a fraction of fraction) is hard to understand, and easy to make mistakes with. If you want, you can simplify it now.

We now have one equation, which involves two variables.

The

lastpiece of information, “the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present“, translates to:T : J = 5 : 2

T / J = 5 / 2

T = (5/2)J

which is the equation you found. Hmm …

we used all the information in the problem, but we only have two equations!That’s OK, since these two equations only use two of my three variables: T and J. So you can solve for T and J. However, the problem asked you to find P!

To recap, we now have two equations in two variables: $$T – 99 = \frac{7}{4}\frac{J}{4}$$ $$T = \frac{5}{2}J$$

While we’re looking at them, let’s go ahead and solve for *T* and *J*:

Clearing fractions, the first equation becomes $$16T – 1584 = 7J$$

Substituting in this from the second equation, we have $$40J – 1584 = 7J$$

Solving, $$J = \frac{1584}{33} = 48$$

Then $$T = \frac{5}{2}J = \frac{5\cdot 48}{2} = 120$$

The ratio of these, as required, is 120:48 = 5:2.

Where does *P* come in?

Well, your mom came to the rescue and solved this dilemma, by writing an expression for P. That’s her last equation:

P=J/4 + (T – 99)

Do you see now how she did that? I think you do.

Recall our definition of *P*: “the cost of the present (in dollars)”. Here we have to think about how things work: If Tom and Jerry split the cost of the present, then its total cost is the sum of what each of them paid. This is a good example of how some equations are implied by knowledge of what is called “the problem domain”. (It can also sometimes be called “common sense”. You didn’t think math had anything to do with that, did you?)

In other words, the Jerry spent \(\displaystyle\frac{J}{4} = \frac{48}{4} = \$12\), and Tom spent \(T – 99 = 120 – 99 = \$21\). These are the answers Eric gave; and if we check, the ratio of these is 21:12 = 7:4. And the total cost of the present is \(12 + 21 = \$33\).

Eric replied:

Ah ok! I think I get it now! Thank you for all your help!

Pondering the problem now, I’ve seen one way it can be solved just by thinking about ratios arithmetically. I started by drawing a picture, which helped largely by making me aware of what we know. Look at the problem again:

Tom and Jerry paid for a friend’s present in the ratio 7:4.

In doing so, Jerry spent 1/4 of his money and Tom had $99 left.

If the ratio of the amount of Tom’s money to the amount of Jerry’s money was 5:2 before they bought the present, how much did the present cost?

Thinking in terms of “parts”, and starting with the last fact we are told, Tom had 5 “parts” to Jerry’s 2 “parts”. If we can find how big a “part” is, we’ve solved the problem.

Jerry spent 1/4 of his money. That’s 1/4 of 2 parts, which means half a part.

The ratio of what they spent was 7:4. That’s 7 “little parts” from Tom and 4 “little parts” from Jerry. So Jerry’s 1/2 “big part” is 4 “little parts”; a “little part” is 1/8 of a “big part.

Now, Tom has $99 left after taking 7 “little parts” from the 5 “big parts” he had. But 5 “big parts” means 5 times 8 “little parts”. Subtracting 7 “little parts” from those 40, the $99 he has left is 33 “little parts”. So one “little part” is $3, and one “big part” is 8 times that, $24. So Tom started with 5 times that, $120, and Jerry started with 2 times $24, or $48. From there, we just find how much each spent and we have the answer.

What I just did reminds me of the way algebra was done before symbols, just talking through it with words. It also reminds me of the way fractions were handled before fractions were invented, by naming units so the fractions would just be whole numbers of smaller units. (If I wanted, I could have found better names!)

Which do you prefer, algebra, or thinking?

]]>We’ll start with this question from 1998:

Triangles and Law of Sines I am having problems with the Law of Sines. To be specific, I don't knowhow to tell if the information I'm given produces more than one triangle. Solving for one is simple. That I can do. But some of the problems in my book produce two triangles, and all the info they give to solve the problem is like 2 sides and an angle or 2 angles and a side. For example: Angle A = 37 degrees Side C = 40 Side A = 28 The back of my book gives two solutions for this this problem. How do you solve it?

Problems with “2 angles and a side” (ASA or AAS) will always have one solution (unless the angles are too large); those with “2 sides and an angle” always have one solution if they are SAS. But there are cases of SSA that have no answer, or one, or two. How do you know which you have? For the moment, we’ll just examine the specific example given here.

Here is what we might draw as a (fairly accurate) sketch of this triangle:

Doctor Scott answered:

Hi Heathe! Using the law of sines you get: 28 40 -------- = ------- sin 37 sin C or 28 sin C = 40 sin 37 40 sin 37 sin C = --------- 28 sin C = 0.859735747

Here we have used the Law of Sines in the form $$\frac{a}{\sin(A)} = \frac{c}{\sin(C)}$$ and solved for the sine of angle C. But, as we saw last time in working with the SSS and SAS cases, there are two angles with the same sine. That will be an issue here, as well:

To solve this equation, we use the inverse sine function. InvSin(0.859..) = 59.28 degrees. But recall that the sin of an angle and the sin of its supplement are equal. That is, sin x = sin (180-x). So, if the sin of 59.29 is 0.8597, then so is sin (180-59.29) or sin (120.71).

The triangle I drew above is the one most of us would expect, and the one we find first: the acute triangle, with a 59.29° angle at C, obtained directly from the inverse sine. The other possible answer, an obtuse triangle with a 120.71° angle at C, looks like this:

It still has the required angle and two sides, but side *a* has been flipped around.

So to find both solutions, we just have to remember to take the supplement of the inverse sine.

But sometimes, as we’ll be seeing, this alternative triangle might not actually exist; we need to check it:

So another possible measure for C is 120.71.We need to see if it can work. Since angle A is 37, keeping in mind that the angles of a triangle add up to 180, we can have C = 59.29 and A = 37 (and still have some angle measure left over for B) OR C = 120.71 and A = 37 and still have some angle measure left over for B.

If the sum of A and C had been greater than 180°, angle B would have turned out negative, and been rejected.

That's why there are 2 solutions in the back of your book. Now, we have tofind all of the other parts of the triangle for BOTH possible angle Bs.

Finishing up, we have:

- for the acute case, \(A = 37°\), \(C = 59.29°\), and \(B = 180 – (37 + 59.29) = 83.71°\); then \(\displaystyle b = \frac{a\sin(B)}{\sin(A)} = \frac{28\sin(83.71°)}{\sin(37°)} = 46.25\);
- for the obtuse case, \(A = 37°\), \(C = 120.71°\), and \(B = 180 – (37 + 120.71) = 22.29°\); then \(\displaystyle b = \frac{a\sin(B)}{\sin(A)} = \frac{28\sin(22.29°)}{\sin(37°)} = 17.65\).

Last time, there was a mention of using the Law of Cosines instead of the Law of Sines for SSA. Since I can’t find any examples of that in our archive, let’s apply that technique to this problem.

Recall the problem:

The Law of Cosines, applied to this triangle, requires writing the unknown side *b*, as a variable: $$a^2 = b^2 + c^2 – 2bc\cos(A)$$ becomes $$28^2 = b^2 + 40^2 – 2(40)b\cos(37°)$$ We can rearrange this to $$b^2 – (80\cos(37°))b + 816 = 0$$

When we solve this by the quadratic formula, we get $$b = \frac{80\cos(37°) \pm \sqrt{(80\cos(37°))^2 -4(816)}}{2} = \frac{63.89\pm28.60}{2} = 46.25, 17.65$$

These are the same answers we got before. As was said last time, this is a little more work than using the Law of Sines, but the formula reminds us that there are two solutions, each of which we pursue to find the other angles.

To observe other things that can happen, consider this similar question from 2003:

The Ambiguous Case I do not understand how to use the ambiguous caseto determine the number of triangles that can be constructed. Is there a simple way to answer the following question: How many triangles can be constructed if, for example, a=4, A=30, and c=12? Or a=9, b=12, and A=35? I am confused about how to do this. Thank you, Les

Both examples are SSA, and we’ll look at those and yet another here.

Doctor Rick answered:

Hi, Les. We can start by applying the law of sines. In your first example, we get sin(C)/c = sin(A)/a sin(C)/12 = sin(30)/4 sin(C) = sin(30)*12/4 sin(C) = 0.5*12/4 sin(C) = 1.5 You know that the sine of any angle is between -1 and 1. There is no angle whose sine is 1.5, therefore there are no solutions in this case.

Here is what this triangle looks like if we try to draw it:

I call this the T Rex case – just imagine those tiny forelegs trying to pick up something it dropped! No matter what direction the 4-unit leg tries to move, it can’t reach the ground and make a vertex C.

And how do we recognize this case? When we find that the required sine is not in the domain of the inverse sine, so we can’t do the final step.

He skipped over the second example, which is the kind we’ve seen above; so let’s do it now:

We have *a* = 9, *b* = 12, and *A* = 35°. The Law of Sines tells us that $$\frac{\sin(B)}{12} = \frac{\sin(35°)}{9}$$ so that $$\sin(B) = \frac{12\sin(35°)}{9} = 0.764769$$ This is a valid sine, so the two possible values of B are \(\arcsin(0.764769) = 49.89°\) and \(180° – 49.89° = 130.11°\). Then,

- for the acute case, \(A = 35°\), \(B = 49.89°\), and \(C = 180 – (35 + 49.89) = 95.11°\); then \(\displaystyle c = \frac{a\sin(C)}{\sin(A)} = \frac{9\sin(95.11)}{\sin(35)} = 15.63\);
- for the obtuse case, \(A = 35°\), \(B = 130.11°\), and \(C = 180 – (35 + 130.11) = 14.89°\); then \(\displaystyle c = \frac{a\sin(C)}{\sin(A)} = \frac{9\sin(14.89)}{\sin(35)} = 4.03\).

The two triangles look like this:

But those aren’t the only things that can happen!

Let's consider another example: B = 50 deg, b = 12, c = 10. In this case sin(C)/c = sin(B)/b sin(C)/10 = sin(50)/12 sin(C) = sin(50)*10/12 sin(C) = 0.638370 C = 39.67 degrees Is this the only solution? Not necessarily, because there is another angle whose sine is 0.638370, namely, 180 - 39.67 = 140.33 degrees. Can we have a triangle with angle B = 50 deg and angle C = 140.33 deg? No, because B+C = 190.33 deg, which is more than the sum of all three angles of any triangle (180 deg). Thus this case is not ambiguous: there is exactly one triangle that satisfies the conditions. The one solution has angle C = 39.67 degrees.

The work here should be familiar by now. But this time, when we check the obtuse solution with C = 140.33°, we find that the resulting triangle doesn’t exist. What does the triangle look like? Here it is:

When we flip side *a* to the other possible position, it is on the wrong side of A, so that the angle opposite the 12 would really be 130°, not 50°. And the angle at the new C would in fact be acute. So the second triangle is a good try, but it fails. As a result, we have only one solution this time.

(Incidentally, the SSA case is often called the “ambiguous case” because of the *possibility* of more than one answer; I tend to think of only the specific situation in which there *are* two solutions as truly ambiguous, and Doctor Rick has used the word in that sense here.)

If the law of sines gives a sine of the missing angle that is less than 1, AND both angles with this sine (the arcsine of the angle and 180 minus the arcsine) are less than 180 minus the known angle, then there aretwo trianglesthat satisfy the conditions.

This is like Les’ second problem, which we looked at above.

Here are some more thoughts: You canrecognizewhether an SSA specification of a triangle has0, 1 or 2 solutionswithout going through the law of sines. Consider the geometry of the triangle. If we have a base (b) of known length and a known angle A, the third vertex (B) must lie along a ray: / / / B / / D / \ / \ B' / \ \a / \ \ / \ \ \ / a\ \ \ / \ \\ /______________________\ A b C We know the length of the side a, and we want to find a point (or points) B along the ray such that BC = a.

Here’s a better picture, which, as you’ll see, is in a different orientation than mine:

Consider the perpendicular CD from C to the ray. Its length is b*sin(A), and it is the shortest distance from C to the ray. If a < b*sin(A), then we know that no point on the ray will be a distance a from C; all points are farther than this. Thus the condition for no solution is a < b*sin(A) In the figure, I show the ambiguous case: both CB and CB' have length a. The triangle BCB' is isosceles, and CD is its altitude, so BD = B'D. This is the ambiguous case: both ABC and AB'C satisfy the conditions.

What if \(a = b \sin(A)\)? Then there is only one triangle, namely the right triangle ACD. This is a very special case.

If we increase length a, then point B moves out along the ray, but B' moves in toward A. When a = b, point B' is coincident with point A. If a > b, we no longer have an ambiguous case: only triangle ABC satisfies the conditions, because B' is in effect pushed off the ray, onto the ray in the opposite direction. Thus, the three cases can be distinguished easily: a < b*sin(A) no solutions b*sin(A) < a < b two solutions b < a one solution In my figure, A is an acute angle.What happens if A is obtuse?It's easy to see that there is no ambiguity. If a > b, there is one solution; if a < b, there are no solutions.

Here are the two cases with an obtuse angle. First, with one solution, because *a* > *b*:

Then, with no solutions, because *a* ≤ *b*:

Now let’s look at what the FAQ says about SSA:

Case II: You are given two sides and the angle opposite one of them, say a, c, and A.Subcase A: a < c sin(A). There is no solution.Subcase B: a = c sin(A). There is one solution: C = π/2, B = π/2 - A, b = c cos(A).Subcase C: c > a > c sin(A). There are two solutions. Use the Law of Sines to find sin(C) = c sin(A)/a < 1. There are two angles C and C' = π - C having that sine, one acute and one obtuse. Then compute B = π - A - C and B' = π - A - C'. Now use the Law of Sines again to find b = a sin(B)/sin(A) and b' = a sin(B')/sin(A). The solutions are (a,b,c,A,B,C) and (a,b',c,A,B',C').Subcase D: a >= c. There is one solution. Use the Law of Sines to find sin(C) = c sin(A)/a <= 1. Then angle C must be acute, so it can be found uniquely from sin(C). Then compute B = π - A - C. Now use the Law of Sines again to find b = a sin(B)/sin(A).

Subcase A is our case (a) above.

Subcase B is the special case where the one solution is a right triangle.

Subcase C is the usual “ambiguous” case with two solutions, our case (b).

Subcase D is our case (c) with one solution.

These don’t include the obtuse cases.

Here is a question from 2002 that is more than just ambiguous:

Different Answers with Sine Rule and Cosine Rule This is really strange. I found this problem in a textbook. A triangle ABC has measurements AB = 8.2cm, BC = 9.4cm and AC = 12.8cm and angle A = 47 degrees. Find angle B. Method 1 - Using the Sine Rule; sin B = AC (sin A)/BC gives B as 84.4 degrees. Method 2 - Using the Cosine Rule; AC^2 = BC^2 + AB^2 - 2(BC)(AB) cos B gives B as 93.1 degrees, which incidentally is the right answer. Why does using the Sine Rule here fail to give the right answer? Thanks for any help you can offer me.

Using the standard labeling, we have *c* = 8.2, *a* = 9.4, *b* = 12.8, and *A* = 47°.

Doctor Rick answered this, too, identifying three different issues that contribute to the confusion:

Hi, Thomas. It's strange, isn't it?There are several things going on here at the same time, making it a bit hard to untangle. First, the triangle isoverspecifiedto begin with. Given just the lengths of the sides, you can use the law of cosines to find angle A. It turns out to be, not 47 degrees exactly, but 47.165 degrees.

Had you noticed that we are given not three, but **four** facts about the triangle? (“Overspecified” means that we are given more information than we need.) It could be called SSSA, and the SSS and the SSA or SAS are at war with one another. As a result, there is really no triangle that satisfies them all! It appears that, as he suggests, the 47 degree angle was given as an approximation, so it can’t really be used to find the answer. On the other hand, in principle you could use any three of the four givens and construct a (different) triangle you might call the “right” one!

The fact that only one angle is requested makes it easy not to check the answer; checking that the three angles add up to 180° would reveal the problem. But so does trying two methods, as Thomas did.

Here is the work for finding the “actual” value of A, assuming the three sides are exact: $$\cos(A) = \frac{b^2+c^2-a^2}{2bc} = \frac{12.8^2+8.2^2-9.4^2}{2(12.8)(8.2)} = 0.679878$$ so that $$A = \arccos(0.679878) = 47.166°$$

Second, sinceangle B is close to 90 degrees, a small error in the sine of the angle can result in a rather large error in the angle itself. In particular, when I apply the law of sines with 47 degrees for A, I get sin(B) = 0.99588589; but when I use 47.165 degrees for A, I get sin(B) = 0.99857047. That's a small error - but when I take the inverse sine of each value, I get 84.80 (I don't know why this is slightly different from your answer) and 86.936 degrees respectively.

Here is the work for these two calculations. First, using A = 47°: $$\sin(B) = \frac{b\sin(A)}{a} = \frac{12.8\sin(47)}{9.4} = 0.99588589…$$ so $$B = \arcsin(0.99588589) = 84.80°$$

Next, using A = 47.166°: $$\sin(B) = \frac{b\sin(A)}{a} = \frac{12.8\sin(47.166)}{9.4} = 0.99857231…$$ so $$B = \arcsin(0.99857231) = 86.938°$$

(I used unrounded values in my work.) In any case, an error of 1/8 of a degree in A led to an error of over 2 degrees in B.

Now look at the corrected answer of 86.936 degrees using the law of sines. Compare it with the correct answer found using the law of cosines: 93.064 degrees.Each is 3.064 degrees away from a right angle - but in opposite directions!Does this give you an idea of what's going on? When you use the law of cosines to find angle B, you are making use of the three sides of the triangle. Remember the SSS congruency theorem:three sides determine the shape of a triangle uniquely. This is why we are able to find a unique measure for angle B. When you use the law of sines, you are making use of two sides and an angle. Notice how these are related: the angle is NOT between the sides. In terms of congruency theorems, we need an SSA theorem - but there isn't one.The triangle is NOT uniquely determined by these quantities.

SSA gives two possible angles, as we’ve seen before; here the correct answer is the obtuse one, not the acute one Thomas found. We are back into the realm of the ambiguity of SSA. But this time, the given sides let us choose which answer is “correct”.

Remember that we found that the sine of angle B is 0.99857047. We can't just take the inverse sine of this number: that is ONE angle with this sine, but 180 degrees minus this angle is another solution. The latter solution turns out to be correct in this case. You see, there are two triangles that have the same angle A and sides AC and BC.The law of sines, applied without thinking, gave one of these triangles, but it was the wrong one.You must be careful in taking the inverse sine: note both solutions in the range 0 to 180 degrees, and check whether each leads to a valid solution of the triangle. If both do, you need more information (such as side AB) to tell which is correct.

Here is the actual triangle, based on the side lengths:

This is not your typical problem; it was given in an inherently inconsistent form. This is not good for a classroom exercise, but is not uncommon in real life! There, we typically have too much information, and some of it will be inaccurate. We have to:

- choose which data are most reliable (in this case, the sides, which were measured more carefully);
- avoid calculations that amplify error (such as inverse sines of ratios close to 1, which are similar to trying to intersect two lines at a shallow angle); and then
- use the other data to check, or to choose between possibilities.

Interestingly, these are exactly the three points Doctor Rick identified in this problem.

]]>First, here is a question from 2006:

When to Use Law of Sines or Law of Cosines When do we use the law of sines as opposed to the law of cosines? I thought that thesine lawis used when we need to find an angle or a side andtwo sides are given but the angle in between them is not given.

Recall that the **Law of Sines** says that $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{C}$$ or, equivalently, that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$ Any pair of fractions can be used, and will involve a pair of angles and their opposite sides, so that if any three of those are known, the other can be found.

The **Law of Cosines** says that $$c^2 = a^2 + b^2 – 2ab\cos C$$ (or the equivalent for angle A or B), involving all three sides and one angle.

Doctor Schwa answered:

Yeah, that's the right idea, Anisha. Thelaw of cosinesrelates3 sides and an angle. So if you know 3 sides, you can use it, or if you know two sides and an angle, you can find the third side. However, because of the form of the equation, if you have an angle that's not between the two sides, you get a quadratic equation in that case--a bit messy. So the law of cosines is best forSSSandSAS, and it's OK forSSAas long as you're willing to solve a quadratic.

We will be looking at the SSS (Side-Side-Side) and SAS (Side-Angle-Side) cases in this post, and the much more interesting SSA next time (including that quadratic equation, which is seldom taught, in my experience). We can represent these cases as:

Thelaw of sinesrelatestwo sides and the angles opposite them. So any time you have two angles (and then can easily figure out the third), it's easy to use the law of sines:ASAorAAS. In theSSAcase, you can use the law of sines, but you have to remember that when you get something like sin(alpha) = 0.7 or whatever,there might be two possible triangles. You have to check alpha = 44.427 degrees or so, but it's also possible that alpha = 135.573 degrees or so.

The ASA (Angle-Side-Angle) and AAS (Angle-Side-Angle) cases use essentially the same method. We will be examining these momentarily.

These cases are:

Note that SSA has been mentioned twice. It is usually taught as an application of the Law of Sines; but can also be solved with the Law of Cosines. Each method involves some subtlety.

So in the SSA case, the law of sines is easier, but you have to remember to check for that second possibility; if you use the law of cosines, you're stuck with the quadratic formula, but the +/- in the quadratic formula immediately reminds you that there are two possibilities that need to be checked. In the other cases it's clear which one to use:SSS or SASmeans the law of cosines,ASA or AASmeans the law of sines.

As we survey these cases, we’ll be seeing how you can decide what to do, if you have not memorized these “rules”.

For the simplest case, consider this question from 1996:

Finding Side Lengths of a Scalene Triangle This following question was on a very good university entrance exam in Brazil, in 1993. It states that: "Two observers on points A and B of a national park see a beginning fire on point C. Knowing that theangles CAB=45 degrees, ABC=105 degreesand that the distance between pointsA and B is 15 kilometers,determine the distancesbetween B and C, and between A and C." Although there was no illustration in the original question, one is roughly drawn below: B /^\<this angle measures 105 degrees 15 km> / \ / \ / \ / \ A /_ _ _ _ _ _ _ _ _\C this is 45 degrees this is, consequently, 30 degrees

This is an ASA triangle; here is a more accurate version:

Doctor Pete answered, initially using special reasoning rather than a standard method (perhaps because the question didn’t mention trigonometry):

The fact that angle BAC is 45 degrees and ACB is 30 degrees was very suggestive to me, so I drew the perpendicular from point B to side AC, which meets at point D. Then BD = AD = AB/sqrt(2) = 15/sqrt(2), since triangle ABD is 45-45-90 and thus isosceles. Also, triangle BCD is 30-60-90, so BC = 2BD = 30/sqrt(2), and CD = sqrt(3)*BD = 15*sqrt(3/2). Therefore the lengths we wish to find are: BC = 15*sqrt(2), AC = AD+CD = 15/sqrt(2)+15*sqrt(3/2) = 15(1+sqrt(3))/sqrt(2).

This involved merely splitting the triangle into a 45-45-90 triangle and a 30-60-90 triangle:

But that was too easy. The general method is to use the Law of Sines, which applies whenever the given facts include an angle and its opposite side. Because the givens do not include any side and the opposite angle, we first need to have found angle C by subtraction, as we did above. Then:

Alternatively, you could use the Law of Sines, which states sin(A) sin(B) sin(C) ------ = ------ = ------ , a b c where A, B, C are angles and a, b, c are the lengths of the sides they subtend (are opposite to). So side AB is "c" in the above equation. sin(A) and sin(C) are easy to find; they are 1/sqrt(2) and 1/2, respectively.

That is, in our triangle we have $$\frac{\sin 45°}{a} = \frac{\sin 30°}{15},$$ using the fact that angle C is 30°. Using known exact sines of the two special angles, we have $$\frac{1/\sqrt{2}}{a} = \frac{1/2}{15}.$$ Doctor Pete then used an angle-sum identity to find the sine of the other angle, B:

sin(B) = sin(105) = sin(45+60) = sin(45)cos(60)+cos(45)sin(60) = (1/sqrt(2))*(sqrt(3)/2)+(1/sqrt(2))*(1/2) = (1+sqrt(3))/(2*sqrt(2)). So we have 1/sqrt(2) (1+sqrt(3))/(2*sqrt(2)) 1/2 --------- = ----------------------- = --- , a b 15 so b = 15(1+sqrt(3))/sqrt(2) = AC, and a = 15*sqrt(2) = BC, which agrees with our previous results.

If we didn’t want or need exact values, we could just use decimal values for the sines: $$\frac{\sin 45°}{a} = \frac{\sin 105°}{b} = \frac{\sin 30°}{C}$$ $$\frac{0.7071}{a} = \frac{0.9659}{b} = \frac{0.5}{15}$$ from which we determine that $$a = \frac{15\cdot 0.7071}{0.5} = 21.213$$ and $$b = \frac{15\cdot 0.9659}{0.5} = 28.978$$ These agree with the exact values above.

If we had been given an AAS triangle, which includes an angle and the side opposite it, we would have done exactly the same work; we would not need to find the third angle to find the first missing side, but would have needed it for the last step.

The FAQ says this about the ASA and AAS cases:

Case I: You are given any two angles and one side c. The third angle is determined from A + B + C = π. Now the Law of Sines can be used to find b = c sin(B)/sin(C) and a = c sin(A)/sin(C).

Note that angles are assumed to be in radians here. Quite often triangles are solved in degrees, in which case you can just replace π with 180.

Next, here is a 1997 question:

Using the Law of Cosines In triangle ABC, side a = 30m, side b = 36m, and side c = 10m. The question asks to find the size of the angle between sides b and c. I think the question has something to do with advanced trigonometry, but I am unsure of how to go about it.

This is an SSS problem, where we are given the three sides; since there are no given angles, we can’t use the Law of Sines:

Doctor Anthony replied, first summarizing the two main tools:

The convention is to use lower case letters a, b, c, to represent the sides of the triangle, and to use upper case letters A, B, C, to represent the angles, with side a opposite angle A, side b opposite angle B, and side c opposite angle C. There are two principal formulae for solving triangles of any shape (not just right-angled triangles). Which one you use depends on what information you are given. You will find these formulae proved in any textbook on elementary trigonometry. Cosine Formula --------------- a^2 = b^2 + c^2 - 2bc*cos(A) or b^2 = c^2 + a^2 - 2ca*cos(B) or c^2 = a^2 + b^2 - 2ab*cos(C) This is used if you are given three sides or if you are given two sides and the included angle. ("Included" means the angle between the two given sides). Sine Formula ------------- a b c -------- = -------- = -------- sin(A) sin(B) sin(C) This is used if you are given two angles and a side, or two sides and a non-included angle.

For this problem, we need to find an angle; the Law of Sines requires either knowing two angles, or having one and needing another, so it can’t be used. Instead, we solve the Law of Cosines for the angle:

In the problem you stated, we are required to find the angle between sides b and c. This is angle A. Since we are given three sides, we use the cosine formula. a^2 = b^2 + c^2 - 2bc*cos(A) 2bc*cos(A) = b^2 + c^2 - a^2 b^2 + c^2 - a^2 cos(A) = ------------------ 2bc

This formula is reasonably memorable, if you need to use it a lot: the sum of the squares of the adjacent sides, minus the square of the opposite side, divided by twice the product of the adjacent sides.

Putting a = 30, b = 36, c = 10 we have: 1296 + 100 - 900 cos(A) = ---------------- 720 496 31 cos(A) = ----- = ----- = 0.68888 720 45 A = 46.458 degrees = 46 deg, 27 min, 28 secs

Here we have used the inverse cosine (or arccosine) function; our angle has a positive cosine, so it is acute; if it were negative, we would get an obtuse angle.

We could do the same thing to find each angle; but we usually don’t:

If you require the other two angles, it is quicker now to use thesine formula. Also, remember that the three angles add up to 180 degrees, so if you have two angles, you can find the third by subtraction from 180 degrees.

Let’s carry this out, which will reveal some risks to be aware of.

Suppose we try to find angle B next. (As we’ll see, this turns out not to be the best idea!)

To find angle B, we would use \(\displaystyle\frac{\sin B}{b} = \frac{\sin A}{a}\), that is, \(\displaystyle\frac{\sin B}{36} = \frac{\sin 46.458°}{30}\), and \(\displaystyle\sin B = 36\cdot\frac{\sin 46.458°}{30} = 0.86984\).

Here’s the trouble: We are likely to simply say \(B = \arcsin(0.86984) = 60.44°\). But there is another angle whose sine is the same, namely the supplement of this angle, \(180° – 60.44° = 119.56°\). How do we know which it is? Because I drew my picture somewhat accurately, we can see that this angle is actually obtuse, so the latter answer is correct.

(I’m choosing to use the notation \(\arcsin(x)\) for the inverse sine; it is also written as \(\sin^{-1}(x)\), especially on calculators.)

The better way is to make sure that we never use the arcsin function to find an angle that *might* be obtuse.

In any triangle, the largest angle is opposite the longest side; and if a triangle has an obtuse angle, that must be the largest angle. Since B is opposite the longest side, it is the only angle that might be obtuse, and just in case, we should not be looking for it at this point. We should either have solved for B first (using the Law of Cosines, which would explicitly tell us if it was obtuse), or leave it for last.

So let’s find angle C this time, by the same method: \(\displaystyle\frac{\sin C}{c} = \frac{\sin A}{a}\), so, \(\displaystyle\frac{\sin C}{10} = \frac{\sin 46.458°}{30}\), and \(\displaystyle\sin C = 10\cdot\frac{\sin 46.458°}{30} = 0.24162\). Therefore \(C = \arcsin(0.24162) = 13.98°\). From this, we can find that \(B = 180° – (A + C) = 180° – 46.458° – 13.98° = 119.56°\).

The FAQ says this about the SSS case (unfortunately not mentioning this issue):

Case IV: You are given all three sides. You can use theLaw of Cosinesto find A, then use theLaw of Sinesto compute sin(B) = b sin(A)/a and sin(C) = c sin(A)/a. Alternatively, you can find r = sqrt[(s-a)(s-b)(s-c)/s], and use tan(A/2) = r/(s-a) to find A/2, and hence A, and similarly for B and C. Alternatively, you can use sin(A/2) = sqrt[(s-b)(s-c)/(bc)] to find A/2 (since A/2 < π/2), and hence A, and similarly for B and C.

The alternatives use ideas related to Heron’s formula, which we discussed in the post Area of a Triangle: Heron’s Formula I. This could be a whole separate post!

A similar example can be found here, where the Law of Cosines is used for all three angles, avoiding the obtuse-angle issue entirely:

Law of Cosines and Pythagorean Theorem

The same issue can arise in the SAS case. Consider this question from 2002:

Laws of Sines and Cosines Recently my math teacher assigned a problem in which the objective was to use the laws of sines and cosines. Here are the specifics of triangle ABC: Angle A is 23 degrees, the side opposite is x. Angle B is y degrees, and the side opposite it is 7. Angle C is z degrees, and the side opposite it is 5. I set up the problem using thelaw of cosines, and found that x = 3.0919. I then set up the rest of the problem usinglaw of sines. However, when I figured the answer, the sine of 23 divided by 3.0919 equals the sine of y over seven, the answer was approximately 62 degrees. I also found that z = approximately 38.168.Obviously these three angles didn't add up to 180 degrees.I figured the problem using cosines, and found that all the answers were the same except the measure of angle y. By using the law of cosines, I discovered that angle y = 118 degrees, approximately. When I figure using the law of cosines, all the angles in the triangle add up to 180 degrees. I noticed that the answer I found for the measure of angle y using the law of sines was the supplement of the actual measure of angle y, using the law of cosines. Assuming that the results are consistent in all problems,why, if an angle is obtuse, does the law of sines find the supplement of the answer?If this does not occur in all obtuse triangles, then why this one? I am really confused, and rapidly losing faith in the laws.

Here is the triangle:

Perhaps you can already see what is happening here! Let’s do the work Alison has described, to get us up to speed:

The Law of Cosines gives \(a^2 = b^2 + c^2 – 2bc\cos A = 7^2 + 5^2 – 2(7)(5)\cos 23° = 9.565\), so \(a = \sqrt{9.565} = 3.0927\). (Presumably Alison rounded somewhere, or copied a number wrong.)

To find angle B (*y*, the second in the list), we use the Law of Sines: \(\displaystyle\frac{\sin B}{b} = \frac{\sin A}{a}\), so, \(\displaystyle\frac{\sin B}{7} = \frac{\sin 23°}{3.0927}\), and \(\displaystyle\sin B = 7\cdot\frac{\sin 23°}{3.0927} = 0.88438\). Therefore (she thinks) \(B = \arcsin(0.88438) = 62.175°\).

She found angle C the same way, and then checked by adding the angles, finding that the sum is not 180°. (Notice that if we just calculated C by subtraction, as we have been doing, then we would miss the conflict! Doing things the hard way and leaving the easy way as a check is a very wise thing to do, Alison!)

Looking at my somewhat accurate picture, we can see the problem: angle B is obtuse.

I answered:

Hi, Alison. Just think about the nature of sines: the sine of, say, 70 degrees, is the same as the sine of 110 degrees (its supplement). Do you see why? Thenany method that gives you the sine of an angle doesn't distinguish between that angle and its supplement; you have two choices for the angle, and have to use other information to decide which is correct. This is something like solving x^2 = 4 by taking the square root; if you just write x = sqrt(4), you have missed one of the two solutions. So the law of sines didn't give you the acute angle; you just forgot to consider the obtuse angle.

Here, it turns out that angle B is \(180° – 62.175° = 117.825°\). And the easy way to find it would have been to find angle C first, which is known to be acute.

Our Dr. Math Trigonometry Formulas FAQ includes instructions for solving triangles: http://mathforum.org/dr.math/faq/formulas/faq.trig.html#solveoblitri This is case III (SAS), and you are told to do just what you did to find x, but then touse the law of sines only for the angles opposite the two smallest sides, which have to be acute. Then you can find the third side by subtracting from 180 degrees, avoiding the need to decide whether to use the acute or obtuse angle with the given sine. I hope that restores your faith!

If the given angle were obtuse, we could proceed without worry; if not, we find the angle opposite the larger given side first.

Here is the FAQ’s entire comment on SAS:

Case III: You are given two sides and the included angle, say a, b, and C. You can compute the third side c by using the Law of Cosines. Then the Law of Sines can be used to find the sines of the other two angles sin(A) = a sin(C)/c and sin(B) = b sin(C)/c. The angles opposite the two shortest sides are then acute, and uniquely determined from their sines, andthe third, largest angleis found from A + B + C = . Alternatively, you can use the Law of Tangents. You know that (A+B)/2 = (-C)/2, which is easily computable. Then by the Law of Tangents, tan[(A-B)/2] = cot(C/2) (a-b)/(a+b), so you can find (A-B)/2 uniquely. Then A = (-C)/2 + (A-B)/2, and B = (-C)/2 - (A-B)/2. Then c = a sin(C)/sin(A).

I’m not sure I’ve ever used the Law of Tangents …

For another example of SAS (together with SSA, if you don’t want to wait for next time), see:

Finding the Third Side

Next time, we’ll dig into the SSA case, which is by far the most complicated; it is the infamous Ambiguous Case.

]]>First, here is a question we looked at last time asking about both the Law of Sines and the Law of Cosines; this time we’ll see the answer to the latter part:

Derivation of Law of Sines and Cosines How do you derive the law of sines and the law of cosines?

Doctor Pete answered:

Generally, there are several ways to prove the Law of Sines and the Law of Cosines, but I will provide one of each here: Let ABC be a triangle with angles A, B, C and sides a, b, c, such that angle A subtends side a, etc. ...

Theorem (Law of Cosines). c^2 = a^2 + b^2 - 2ab Cos[C]. Proof. Place triangle ABC on a Cartesian coordinate system such that angle C is at the origin and length a lies on the x-axis. Then length b is on the other ray from the origin. We can easily identify the coordinates of two of the vertices: Vertex C lies at (0,0), and vertex B lies at (a,0). We must identify vertex A, which lies somewhere in quadrant I or II (since angle C < 180 degrees). But thex-coordinate of A is b Cos[C], as can be seen by drawing the perpendicular from A to the x-axis. Similarly, they-coordinate is b Sin[C]. Hence A lies at (b Cos[C], b Sin[C]).

Therefore, the length of c is given by the distance between A and B, which is by the distance formula c = Sqrt[(a - b Cos[C])^2 + (b Sin[C])^2] or upon squaring both sides, c^2 = (a - b Cos[C])^2 + (b Sin[C])^2 = a^2 - 2ab Cos[C] + b^2 Cos[C]^2 + b^2 Sin[C]^2 = a^2 + b^2 (Cos[C]^2 + Sin[C]^2) - 2ab Cos[C]. But Cos[C]^2 + Sin[C]^2 = 1 for any angle C, and therefore c^2 = a^2 + b^2 - 2ab Cos[C].

So the work is mostly algebra, with a trig identity thrown in. In fact, we used the Pythagorean Theorem at least twice, first in the form of the distance formula, and again in the form of the Pythagorean identity, \(\sin^2 \theta + \cos^2 \theta = 1\).

I hope that was understandable. To really get a feel for the proofs, draw out the diagrams. Clearly, the Law of Cosines is a more difficult proof, but it is really ageneralization of the Pythagorean Theorem, since in the right-angle case (C = 90), Cos[C] = 0 and we obtain c^2 = a^2 + b^2. This is why we often see the Law of Cosines written as c^2 = ... instead of the equally valid form a^2 = b^2 + c^2 - 2bc Cos[A].

Applying the Law of Cosines to each of the three angles, we have the three forms

$$a^2 = b^2 + c^2 – 2bc\cos(A)$$

$$b^2 = a^2 + c^2 – 2ac\cos(B)$$

$$c^2 = a^2 + b^2 – 2ab\cos(C)$$

A virtually identical proof is found in this page we also looked at last time:

Proving Laws of Sines, Cosines

The next question was from a student who just guessed that there should be a way to modify the Pythagorean Theorem to work with non-right triangles; that is just what the Law of Cosines is. But since Brooke apparently does not know trigonometry yet, a mostly geometrical answer seemed appropriate.

Deriving the Law of Cosines I wonder if the Pythagorean Theorem will work with a non-right triangle and how to go about proving it.

Doctor Rob answered:

This is a good question, and it shows good thinking. The Pythagorean Theorem will not work for triangles that are not right triangles. When you have a triangle which is not right,the closest analogue is something called the Law of Cosines. If the sides of a triangle are labeled a, b, and c, and the angles opposite them A, B, and C, respectively, then: c^2 = a^2 + b^2 - 2*a*b*cos(C) Here "cos(C)" means the cosine of the angle C. If the triangle is a right triangle with right angle C, then the cosine of C, that is, the cosine of 90 degrees, is zero, so you get c^2 = a^2 + b^2, the Pythagorean Theorem.

So the Pythagorean Theorem can be seen as a special case of the Law of Cosines.

If the angle is other than 90 degrees, however, the cosine of C is not zero. For example, if the angle C is 60 degrees, its cosine is 1/2, and you get the equation c^2 = a^2 +b^2 - a*b. If C is 120 degrees, its cosine is -1/2, and you get the equation c^2 = a^2 + b^2 + a*b.

You may find it interesting to see what happens when angle C is 0° or 180°! These are not literally triangles (they can be called degenerate triangles), but the formula still works: it becomes mere addition or subtraction of lengths.

Now he gives an algebraic proof similar to the one above, but starting with geometry rather than coordinates, and avoiding trigonometry until the last step:

You will learn about cosines and prove the Law of Cosines when you study trigonometry. The proof depends on the Pythagorean Theorem, strangely enough! Draw triangle ABC with sides a, b, and c, as above. Drop a perpendicular from A to BC, meeting it at point P. Let the length AP be y, and the length CP be x. Then BP = a-x. Then apply the Pythagorean Theorem to the two triangles ACP and ABP: CP^2 + AP^2 = AC^2 or x^2 + y^2 = b^2 BP^2 + AP^2 = AB^2 or (a-x)^2 + y^2 = c^2

(I’ve swapped the names of *x* and *y* from the original, to increase the similarity to our coordinate proof above.)

Now subtract the first equation from the second, to eliminate y: (a-x)^2 - x^2 = c^2 - b^2 a^2 - 2*a*x + x^2 - x^2 = c^2 - b^2 a^2 - 2*a*x = c^2 - b^2 c^2 = a^2 + b^2 - 2*a*b*(x/b)

This is the non-trigonometric version of the Law of Cosines.

Now x/b is just the cosine of angle C (if you don't know this, trust me), so: c^2 = a^2 + b^2 - 2*a*b*cos(C) Observe that C is a right angle if and only if x = 0 and points P and C coincide.

Here is a question from 2006 that was not archived:

Hi, I'm a student at the University of Minnesota Twin Cities Campus, and I'm an Actuarial Science major. I'm writing an article, and it involves stating the proofs of the Law of Sines and the Law of Cosines. I found the proofs very easily by researching on the internet, but I would like to add one possible sentence onwho first proved the Law of Sines and the Law of Cosinesif possible, just to make it a bit more interesting. If you know, please let me know. Thank you very much.

I replied,

The Law of Cosines, at least, can be found in Euclid -- which wasbefore cosines were defined! See this page: http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII12.html For another discussion, see http://www.cut-the-knot.org/pythagoras/cosine2.shtml I don't find that the Law of Sines is found in Euclid, but it can be proved easily from Euclid, and in some form was proved quite early. For more details, see http://en.wikipedia.org/wiki/Law_of_cosines http://en.wikipedia.org/wiki/Law_of_sines The latter, unfortunately, doesn't mention history, as the former does.

The Cut-the-Knot page includes several proofs, as does Wikipedia.

Euclid has two propositions (one applying to an obtuse triangle, the other to acute), because negative numbers were not acceptable then (and the theorems don’t use numbers in the first place, but lengths!). Neither trigonometric functions nor algebraic concepts existed yet, so everything had to be expressed in terms of geometry. This makes for a very interesting perspective on the proof!

Let’s just look at the acute case:

Euclid’s Elements: Proposition II.13

In acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle.

If you never realized how much easier algebraic notation makes things, now you know!

If we label the triangle as in our previous figures, we have this:

The theorem says, in the geometric language Euclid had to use, that:

The square on the side opposite the acute angle [ \(c^2\) ] is less than the sum of the squares on the sides containing the acute angle [ \(a^2 + b^2\) ] by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls [*a*], and the straight line cut off within by the perpendicular towards the acute angle [*x*, so the rectangle is \(2ax\)].

That is, $$c^2 = a^2 + b^2 – 2ax$$

Since \(x = b\cos(C)\), this is exactly the Law of Cosines, without explicit mention of cosines.

I won’t quote the proof, which uses different labels than mine; but putting it in algebraic terms, it amounts to this:

From a previous theorem (Proposition II.7), $$a^2 + x^2 = 2ax + y^2$$

[This amounts to our algebraic fact that \(y^2 = (a – x)^2 = a^2 – 2ax + x^2\).]

Adding \(h^2\) to each side, $$a^2 + x^2 + h^2 = 2ax + y^2 + h^2$$

But from the two right triangles \(\triangle ACD\) and \(\triangle ABD\), \(x^2 + h^2 = b^2\), and \(y^2 + h^2 = c^2\). So our equation becomes $$a^2 + b^2 = 2ax + c^2$$

Rearranging, we have our result: $$c^2 = a^2 + b^2 – 2ax$$

Again, we have a proof that is substantially the same as our others – but this one is more than 2000 years older!

]]>I’m in the middle of discussing the Law of Sines and the Law of Cosines, and in searching for questions about them, I ran across one that stands by itself. A student asks his teacher why his method without trig doesn’t work, and gets three answers from us. They are all similar, but from slightly different perspectives.

Here is the question from the teacher (2001):

Find Angle Given 3 Sides, NOT Using Law of Cosines I teach a Math Analysis course where we cover trigonometry. A student asked me, after taking a test, about a problem whereI had given them 3 side lengths and asked them to find an angle. He said he originally couldn't remember the formula so he added all three sides, took 180 degrees divided by the perimeter to determine thedegree per unit, and then multiplied that result by the length of the side opposite the angle I asked for. Later he remembered the formula and when he reworked the problem the answer differed from his first by over 10 degrees.He asked me why it wouldn't work to use proportions.I can not really come up with a good reason why - his process seemed to make sense to me, but I knowit can't be right or we wouldn't need the law of cosinesto find an angle. What is the fault in his logic?

“Why doesn’t the wrong method work?” can be hard to answer sometimes! We could just say, “That’s not how we do that;” but when the wrong method felt intuitively right to the student, we want to overcome the underlying error.

Let’s first think about the problem they are discussing. Perhaps we are told the lengths are *a* = 2, *b* = 3, and *c* = 4, and we are to find angle A.

The Law of Cosines, which we’ll see next week, is the appropriate tool for the problem. This says that $$a^2 = b^2 + c^2 – 2bc\cos(A)$$ which we can solve for angle A as $$\cos(A) = \frac{b^2 + c^2 – a^2}{2bc} = \frac{3^2 + 4^2 – 2^2}{(2)(3)(4)} = \frac{21}{24} = 0.875$$ so that \(A = \arccos(0.875) = 28.955°\):

The student’s work would be to divide \(\frac{180°}{2+3+4} = \frac{180°}{9} = 20°\) per unit of perimeter; so \(A = 2\times 20° = 40°\), which is more than 10° too large.

So clearly the method is wrong; how to explain it to him? The basic idea is that his work is a counterfeit Law of Sines, without the sines. Why doesn’t it work without the sines?

Doctor Schwa answered first:

Thelaw of sinestells us that the length of each side is proportional to the SINE of the angle opposite it, and your studentassumed that it was proportional to just the angle. In other words, an angle twice as big makes the side opposite it less than twice as big. Your student assumed it would be twice as big.

In our example, angles A and B are, respectively, 28.96° and 46.57°. Their ratio is about 1:1.608, while the opposite sides, 2 and 3, are 1:1.5. Close, but not exact. And angle C is even farther than it “should” be.

Another, maybe more intuitive way to say the same thing is: if you make the angle twice as big, thearc of a circlewould get twice as long. But the length of the side of a triangle, which is like thechord of the circle, would get less than twice as long. The two arcs add up end to end but the two chords make a bent line, and the new (straight) chord is shorter. I hope that helps clear things up!

Here is an example of an arc that has been doubled (from 40° to 80°), but the chord has less than doubled (from 1.4 to 2.63, not 2.8). It is clear that the two shorter chords make a triangle with the longer chord, so the latter is shorter than the sum of the former:

Within a triangle, there is even more going on, as each angle is a different *distance* from the opposite side. But the simple difference between an arc and a chord is enough to see the error.

Seven minutes later, Doctor Rob finished his own answer, focusing on the student’s idea of finding the number of degrees per unit of perimeter:

The problem is thatsome degrees in angle A cut off larger portions of side a than others do. Thus the side lengths are not proportional to the angles. The side lengths are proportional to the SINES OF THE ANGLES, and not the ANGLES THEMSELVES. If they were, the formula corresponding to the Law of Sines would be a/A = b/B = c/C = (a+b+c)/(A+B+C) = P/180, and not a/sin(A) = b/sin(B) = c/sin(C) = 2*R. The former is what the student used. The latter is correct.

He has explicitly shown the contrast between the student’s assumed proportion and the real proportion, the Law of Sines. (You’ll note that he put the latter in the more complete “side to sine” form we discussed last time, which is not taught as often as it should.)

What did he mean by “some degrees cut off larger portions”? Suppose we rotate step by step, one degree at a time, through angle A. Each rotation cuts off a different length of the opposite side:

(I actually rotated AB by 2° at a time, rather than 1°, to make it easier to see.) This shows visually that the length of the side is not proportional to the angle; where the side is more distant, and at a steeper angle, there is more side per degree. There is not a constant number of degrees per unit of side, as there would be for a true proportion.

Doctor Rob continued with a separate thought about the problem itself:

Funny thing, when I saw the subject of your message ["Find angle of triangle given 3 sides and NOT using law of cosines"], I was sure you were going to askhow to calculate the angles from the side lengths without using the Law of Cosines. This is possible, but little known. The trick is tocompute the area two ways. First K = sqrt(s*[s-a]*[s-b]*[s-c]), where s = (a+b+c)/2, (Heron's Formula), and then the area also is given by K = (1/2)*a*b*sin(C), sin(C) = 2*K/(a*b). Similarly sin(A) = 2*K/(b*c), sin(B) = 2*K/(c*a). You may want to add this fact to your store of knowledge.

This puts together some of the pieces we used last time to prove the Law of Sines; it makes sense that we can bypass the law by retracing parts of its proof.

Taking my example, we find that for *a* = 2, *b* = 3, *c* = 4, \(\displaystyle s = \frac{2+3+4}{2} = 4.5\), and $$K = \sqrt{s(s-a)(s-b)(s-c)}\\ = \sqrt{(4.5)(4.5-2)(4.5-3)(4.5-4)}\\ = \sqrt{(4.5)(2.5)(1.5)(0.5)} = 2.9047$$ Then $$\sin(A) = \frac{2K}{bc} = \frac{2\cdot2.9047}{3\cdot 4} = 0.48412$$ and therefore $$A = \arcsin(0.48412) = 28.955°$$ This is just what we got above using the Law of Sines.

Another fourteen minutes later, having seen the ideas already given, I gave it my own twist, first restating the essentials:

Hi, Diane. I see you already got two answers, but a third may give a slightly different perspective. There'sno real logic(in the sense of something provable) in his reasoning. The foundation of what he did doesn't even exist: it's meaningless to talk about the"degrees per unit"in a triangle, because theangles are not proportionalto the lengths of the opposite sides. You can only work with ratios where some sort of proportion exists.

I find it very common for students who have learned about proportions to *assume* two quantities are proportional just because that would make their work easy. Before applying proportions, it is necessary to have a *reason to expect that model to be valid*. I don’t think that is taught often enough.

My new perspective was to consider what the student’s method really does:

What he's really finding is something like this. Make a circle whosecircumferenceis the sum of the three lengths a, b, and c: *********** c ***** ***** **** **** A ** ** +---- * ** \ ---------- ** * \ --------- * * \ ---------- * * \ ----+ B * \ / * * \ + / * * \ O / * * \ / * * \ / * b * \ / * ** \ / ** * \ / * ** \ / ** a **** \ / **** ***** \ / ***** ****+****** C

That is, suppose that *a*, *b*, and *c* were not the *side lengths*, but the *arc lengths* along the circumcircle. Then:

The central angles subtended by the arcs will be proportional to the arc lengths, so dividing the 360 degrees by the circumference (the sum of the lengths) gives thecentral angle per unit along the circumference. The central angle subtended by each arc is therefore the product of this ratio and the length of the arc. Theinscribed angle subtending any arc is half the central angle, so angles A, B, and C are the angles he calculated: A = 180a / (a+b+c) B = 180b / (a+b+c) C = 180c / (a+b+c)

He divided the total angle in proportion to the arc lengths.

But this has nothing to do with the triangle with _sides_ a, b, c, except to the extent that our triangle ABC whose _arcs_ have these lengths approximates a similar triangle - which, as you can see, is not a close approximation in general. So, why doesn't it work to use proportions? Becausethe sides are not proportional to the angles. (They are, of course, proportional to the _sines_ of the angles; but you don't know the sum of the sines as you do the sum of the angles, so that doesn't help.)

All of our answers say essentially the same thing: that while angles are proportional to arcs, and sines of angles are proportional to sides, the angles themselves are not proportional to sides. The student really did know this; the lesson is never to assume something you don’t know to be true. What sounded on the surface like logic was really unfounded hope!

]]>Here is a question from 1997:

Derivation of Law of Sines and Cosines How do you derive the law of sines and the law of cosines?

Doctor Pete answered; we’ll only look at the Law of Sines part today:

Generally, there are several ways to prove the Law of Sines and the Law of Cosines, but I will provide one of each here: Let ABC be a triangle with angles A, B, C and sides a, b, c, such that angle A subtends side a, etc. Theorem (Law of Sines). Sin[A]/a = Sin[B]/b = Sin[C]/c. Proof. Draw the perpendicular (altitude) from B to side b, and suppose this has length h. Then note that Sin[C] = h/a, or h = a Sin[C]. Similarly, Sin[A] = h/c, or h = c Sin[A]. Therefore, a Sin[C] = c Sin[A] or Sin[A]/a = Sin[C]/c, since neither a nor c has length 0.

Here is his picture:

$$h=a\sin(C), h = c\sin(A)\Rightarrow c\sin(A) = a\sin(C)\Rightarrow \frac{\sin(A)}{a} = \frac{\sin(C)}{c}$$

The last step was to “un-cross-multiply”, that is, we divided both sides by *ac*.

Clearly, if we were to draw another altitude h', say, from A to a, we would find by a similar argument that Sin[B]/b = Sin[C]/c, and hence the result follows.

$$h’=b\sin(C), h’ = c\sin(B)\Rightarrow b\sin(C) = c\sin(B)\Rightarrow \frac{\sin(B)}{b} = \frac{\sin(C)}{c}$$

Doctor Pete here has used what I call the “sine over side” formulation of the Law, $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$

In the post Circumcircles and the Law of Sines, I showed Doctor Pete’s answer in __Sine of an Angle and Opposite Side__, where he was asked about it effectively in that form, but showed a different proof that actually demonstrated a fuller form (which I call “side over sine”) that also involves the diameter of the circumcircle: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$ We’ll see this form, and the associated proof, again below.

At the end he offered another proof:

Another little tidbit of interest is that the proof of the Law of Sines can be taken in a slightly different direction to show that the area of a triangle is (ab Sin[C])/2 = (bc Sin[A])/2 = (ac Sin[B])/2.

The formulas for area are derived from the familiar formula \(K = \frac{1}{2}bh\), using the formulas above for *h*; for example, \(h = c\sin(A)\), so \(K = \frac{1}{2}b\cdot c\sin(A) = \frac{1}{2}bc\sin(A)\).

But this can be turned into a proof itself: Using the three angles to find the area, we get $$\frac{1}{2}bc\sin(A) = \frac{1}{2}ac\sin(B) = \frac{1}{2}ab\sin(C);$$ and when we multiply each expression by \(\frac{2}{abc}\), we get $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$

For another version of the proof via altitudes, see this answer from 2003:

Proving Laws of Sines, Cosines

A different proof showed up within an answer to a different question in 2001. First, we need to see the original question, which is interesting in itself:

Radius of Circumscribed Circle On your page on Triangle Formulas in the Dr. Math FAQ at http://mathforum.org/dr.math/faq/formulas/faq.triangle.html you give the formula R = abc/4K for theradius of the circle circumscribed about the scalene triangle abc, where K is theareaof that triangle (obtainable, e.g., from Heron's formula). Where (online, or if necessary in the literature) can I find aderivationof R = abc/4K? Or can you provide one yourself? (I recognize that geometry in an essentially ASCII medium is tedious.) I am interested in a derivation because I am an assistant MathCounts coach and would like to have answers for bright kids. Thanks (and thanks for a great Web site!),

In this formula, *K* is the area of the triangle (we use *K* rather than the more natural *A* for area to avoid confusion with the angle *A*); *a*, *b*, and *c* are the side lengths as usual; and *R* is the circumradius (the radius of the circumscribed circle, which contains the three vertices):

Doctor Schwa answered:

Hi Peter, The derivation that I know is a bit above the level a MathCounts kid could appreciate: they don't generally know any trigonometry yet. I start with the area K = 1/2 ab sin C, and then use the law of sines, c/sin C = 2R to get sin C = c/2R, whereupon K = abc/4R. I don't know of any more elementary geometric proofs of this fact.

If you haven’t seen the “side over sine” formulation of the Law of Sines, then you are probably wondering what hat he pulled the formula \(\displaystyle\frac{c}{\sin(C)} = 2R\) out of! We’ll get there. But (accepting that for now) here is the rest of this derivation, which starts with the area formula we saw earlier: $$K = \frac{1}{2}ab\sin(C) = \frac{1}{2}ab\cdot \frac{c}{2R} = \frac{abc}{4R}$$

Doctor Schwa’s statement that he doesn’t know a geometric proof (one that avoids the trigonometry) led me to look for one, and I’ll offer that after we finish this discussion!

Peter mentioned using this formula to find the circumradius given only the lengths of the sides, using Heron’s formula, which we discussed in the post Area of a Triangle: Heron’s Formula I. This formula calculates the area *K* of a triangle from the lengths of its sides; combining that with the formula proved here allows us to find the circumradius. Specifically, this will be $$R = \frac{abc}{4K} = \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}} = \frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}.$$ (Here \(s=\frac{a+b+c}{2}\).)

A year and a half later, a reader, Gary, wrote to ask about that same line:

In your solution for the radius of a circumscribed circle about a triangle, you magically stated thatbecause of the Law of Sines, sin C = c/(2R). I could not derive that from the Law of Sines, but rather from the fact that when radii are drawn to the vertices, the base triangle with side C has an acute angle which is the complement of the vertex angle C. This means that sin C = cos (base angle of lower triangle) which is 1/2(c)/R, and thus the substitution. I would be interested in seeing your derivation from the Law of Sines.

Clearly Gary is familiar with the law in the “sine over side” form we’ve seen so far, which deals only with ratios, and makes no mention of *R*. He offers a different way to obtain the same fact, which as we’ll see really amounts to a proof of the Law of Sines.

Here is his proof:

The green angle at A is the complement of the blue angle at C; I’m not sure how Gary determined that, but I would say that angle AOB, the central angle of arc AB, is twice angle C, so the blue angle at O is congruent to angle C. The cosine of the green angle is \(\frac{c/2}{R}\), and is also the sine of angle C. Therefore $$\sin(C) = \frac{c}{2R}$$ and $$\frac{c}{\sin(C)} = 2R$$ This is essentially identical to the proof I previously mentioned, in the post Circumcircles and the Law of Sines.

Doctor Schwa replied, first just stating the Law of Sines in the side-over-sine form as he knows it (and Gary did not):

Since the law of sines says a / sin A = b / sin B = c / sin C = 2R it's pretty easy to derive! I knowa lot of books leave off the "= 2R" part, and some even go so far as to put the sines on top instead of on the bottom. I think the best way to prove the law of sines is to prove that a / sin A = 2R and then by symmetry the rest follows. So I am pretty insistent on getting that 2R in there.

This makes Gary’s proof another proof of the Law of Sines:

Of course your proof that sin C = c/(2R) isequivalent to proving the law of sines(when you supplement it with the symmetry argument to show that it must also be true for B and A).

Since Gary had not fully stated the details of his proof, Doctor Schwa made his own explicit:

My proof is just like yours: I draw in a radius of the circle, and show that (c/2) / sin C = R, and thus c / sin C = 2R, and then by symmetry, the law of sines is proved. I think this is a good argument, especially for students like mine where sin and cos are introduced as circular functions in the first place: why not use the circle?

Yes, this is essentially the same as what Gary said. Here is an alternative way to get that ratio, a little more directly:

I have drawn in diameter \(AC’\); the angles at \(C\) and \(C’\) are congruent (because they are inscribed angles subtended by the same arc \(AB\)), while \(ABC’\) is a right triangle (because it is inscribed in a semicircle). Consequently, \(\sin(C) = \sin(C’) = \frac{c}{2R}\). The rest follows as before.

I promised a direct geometric proof of the area formula, which we can accomplish by simply incorporating part of the proof of the Law of Sines.

We can use the last figure above, including an altitude:

Observe that \(\triangle BCD\sim\triangle AC’B\) (since they are right triangles with congruent acute angles). Therefore, \(\displaystyle\frac{h}{a} = \frac{c}{2R}\) and \(\displaystyle h = \frac{ac}{2R}\).

Now $$K = \frac{1}{2}bh = \frac{1}{2}b\cdot \frac{ac}{2R} = \frac{abc}{4R}$$ This leads directly to our goal: $$R = \frac{abc}{4K}$$

]]>I enjoy getting questions from young children, as we did here. It forces us to try to express big ideas in simple words (or at least help their parents or teachers do so). A frequent subject of those questions is infinity – they seem fascinated by this concept, perhaps for the same reasons they love dinosaurs! We’ve looked at the subject from a relatively advanced perspective (and will surely do so again), but here we’ll focus on how to talk about it with kids.

Here is a little question that came in last November, that will give us an opening to a literally huge subject.

This question is from a 5 year old:

If you had a bottomless jar of cookies (i.e. a jar with a top, but that went on infinitely in the other direction)…

If you filled that jar with infinite cookies, would the cookies reach the top of the jar?

The question immediately reminded me of past discussions at a higher level; but what can I say here? I gave it a try:

Hi, Colin.

Great question! I love helping kids think deeply. (

Verydeeply!)My first thought is very simple: To my mind,

“fill”, so the answer is yes, by definition.meansto fill to the top

Here’s my idea of a (transparent) cookie jar filled to the top (with the bottom fading off into the distance), and another not quite “full”:

My answer does feel sort of like cheating; but it makes a real point: We have to decide what we mean by “fill”.

Of course, that’s not the intent of the question! So,

whatPresumably, just to put in the number of cookies it can hold, which is infinite. The answer in that case is,doesit mean to fill the jar?it might, or it might not. We can’t know.

This, again, is a serious answer. If you imagine filling the jar *from the bottom*, you can’t even really think about it – there is no bottom! As I’ll explain next, this corresponds to a real issue for which mathematicians even have a big word:

The trouble is that infinity doesn’t behave like real numbers. If a jar were 6 inches deep and I put in 5 inches of cookies, it would be filled to 1 inch below the top: 6 – 5 = 1. But we can’t do that with infinity; ∞ – ∞ is what we call an

indeterminate form, which means the answer depends on how you get to it. It might be that the jar is exactly filled, or the top cookie might be a mile down, or it might be overflowing!

That is, the distance down from the top is the difference between the height of the jar and the height of the pile of cookies, and when both are infinite, there is no one correct answer.

But how do we explain this conclusion? I took a cue from a famous story about “Hilbert’s Hotel” or “Hotel Infinity”, which I’ll refer to later:

To illustrate this, suppose the jar has an infinite number of shelves that can hold one cookie each, and it’s full to the top. Now just move each cookie one shelf down. There is always a shelf below the shelf we’re looking at, so every cookie can be moved down. But now the jar is filled only to one shelf from the top, yet still holds an infinite number of cookies. Keep doing that, and you can have a jar filled to any level you like, but always holding an infinite number of cookies.

Of course, that’s nonsense in the real world; there is no such thing as an actual infinity. It’s also why we can’t treat infinity as a number.

Pushing the whole stack of cookies down doesn’t change the number of cookies, but does change the level. This is why we can’t know what level the cookies will be at!

I closed with comments for the father:

For some similar ideas (well above 5-year-old level), see https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel and the references at the bottom. (The hotel is just a bigger version of your cookie jar.) We also talked a little about infinity minus infinity here:

These last two references deal with the fact that subtraction and similar operations on infinity are “indeterminate forms”, the first in terms of sets (as we’ll be considering below), and the second in terms of limits in calculus.

The Wikipedia article is about the idea discussed next:

Let’s look at a couple references in the *Ask Dr. Math* archive to similar ideas:

First, a question from slightly older kids in 1995:

Question About Infinities From one of our 4th graders: If there are two points in space, A and B, and set 1 equals all the possible paths which lead away from point A (this set is assumed to be infinite) and set 2 equals all the possible paths which lead away from point A which do not pass through point B (this is also assumed to be infinite), if I subtract set 2 from set 1, does this prove that there are no paths which lead away from point A which pass through point B?

This is a far more complex question than our five-year-old’s! But it also involves subtraction of infinities. Here we have an infinite set (all paths) from which we take away an infinite subset (paths not through B), and our experience with subtraction leads to the conclusion that the result should be zero. But clearly the resulting set is not empty. So what is happening?

Doctor Steve answered, starting by pointing out this connection to subtraction:

Great question! It gave us an occasion to take a little side excursion into orders of infinity, even though that really doesn't come into play here. The problem gets clearer if we clarify some language. To figure out what thedifferenceisbetween two setsis not the same as finding thedifference between the number of elementsin the sets. It is common to define the difference between sets A and B as all the elements which are in one but not the other. Note that I wrote difference rather than subtract, because sets are not numbers and you don't treat them exactly the same way. The number of elements in a set is not the set; it is one of the set's characteristics.

For *finite* sets, the size of the difference between two sets is the difference in their sizes. But we need to distinguish the two concepts.

So if we want to know thedifference between set 1 and set 2in your problem, it becomes fairly easy to state:all the paths from point A which pass through B. If we want to know what happens when wesubtract the number of elements in B from the number in A, then we have a separate problem which doesn't tell us anything about the elements themselves, just about the number of them.

We are subtracting set 2 from set 1, one infinite set from another. Are we really subtracting infinity as a number?

You want to know what happens when wesubtract infinity from itself. And it seems as though just about anything could happen when you do that. My friend likes the example ofHilbert's hotel. Imaginea hotel with an infinite number of rooms and a person in each room. Now suppose a thousand people leave (imagine the rooms are numbered 1,2,3,... and the people in the first thousand rooms leave).Would there be any fewer people?Would we have to leave some rooms empty? No, we just reassign everyone left to a new room (in my example we could do it by having everyone go to a room 1000 less than the number they were in). So here we have two infinities. The nature of infinity is that if you take some finite amount away from it, there's still an infinite amount left. Andwe're tempted to say in this case that the difference between these two infinities is 1000. But we could do the same thing with 300 people leaving and then we'd say the difference between the infinities is 300. Maybe you can begin to see thatwe can subtract infinity from itself and make the answer anything we want.

This is the same as my illustration of moving each cookie down one shelf, with the same conclusion, that the difference (the number of shelves at the top that are empty) can be anything.

Well, mathematicians tend not to like what happens to math if the result of an operation is "whatever you want it to be." If you can't define what the result is, we say that it is undefined, rather than let the answer be anything at all. So when you ask us what happens when you subtract infinity from itself we say that operation is not defined in mathematics.

We often say that infinity is not a number. If we chose to call it a number, it would have to be a special kind of number that can’t be part of certain operations, such as subtraction. So we avoid complication by just admitting that it doesn’t behave like a number, so it isn’t.

Another 1995 answer, Infinity Plus One, about a somewhat different question, referred to a very nice version of Hilbert’s Hotel. The site referred to no longer exists, but the story at the core of it can be found here (as linked from the Wikipedia page):

Welcome to the Hotel Infinity, by Nancy Casey

A 1999 question was about a problem inspired by one version or another of this story:

Infinity Hotel Paradox Our math class has a paradox we can't figure out. Can you make heads or tails of it?A hotel has an infinite number of rooms. Each room has already been assigned to a person with the corresponding room number on his or her shirt; i.e. the person with a shirt that says 438 is in room 438, etc.Now an infinite number of buses come, each containing an infinite number of people.They all want a room. Explain how all of these can be accommodated while still maintaining the original premise that each person gets his or her own room. Also, you can only use positive integers, no negatives or irrational numbers.

The answer, from Doctor Rob, involves algebra and other ideas beyond what we are covering here, but it’s there to read if you wish …

Let’s look at a few past questions from kids on the subject.

First, here is a question from fourth graders:

What is the Largest Number? We are a 3/4 class, studying large numbers. We have a question for you.We want to know what the highest number is. Sincerely, Mrs. Stanko's students

Questions about large numbers are very common, and they lead toward the idea of infinity. Doctor Ethan answered this one:

Neat Question! I think the reason that you are having trouble finding the answer is becausethere is no answer. Let's see why. 1,000,000,000 (1 billion) can't be the largest number because 1 billion + 1 is bigger. But that is true for any number you pick. You can pick any big number and I can pick a bigger one just by adding 1 to it.

Even if you couldn’t name the number, I could tell you that it isn’t the largest just by saying “add 1 to it!”

Mathematicians make up a name for the quantity bigger than all the numbers. They call it infinity, but infinity isn't a number, it is just a name for the thing bigger than all numbers.

As we’ve said before, infinity isn’t a number; but it can make sense to call it a “quantity”.

Here’s a similar question, this time from a five-year-old, passed on through a mother who knows the answer but can use some help:

Do Numbers Go On Forever?Do numbers go on forever, or do they stop? What is thelargest number that has a name? P.S. from Mom: What's the best way to explain infinity to a kindergartener?!

There are really three questions here: whether there is a last number, what is the largest named number, and infinity. I started by answering the first question, which is the same as the previous one:

Hi, Christopher and Mom. This is something very special about numbers, which is worth thinking about:numbers are just ideas, not anything you can see or touch, sothere is nothing to stop them from going on forever. We have to use our imaginations and picture what it would be like if we found a "largest number." Suppose I came up to you and said, "The largest number is umpteen quillion." (I made that up!) You would just laugh at me and say, "No it's not! Umpteen quillion and one is bigger!" Do you see how you could make a fool of me if I tried to give you an answer like that?

The fact that numbers are mere ideas is very important. A number can “exist” even if there aren’t that many of anything in the universe, and even if no one has named it. We just know that in principle, we can go beyond any number you can imagine. (This is a simple example of mathematical induction, by the way.)

Now, how about *named* numbers?

Now, it's a little trickier to ask, "What is the largest number _that has a name_?" You could find the largest number name in a dictionary; that is probably "centillion." (Try looking it up.) But mathematicians are not satisfied with mere lists of words. Mathematics instead works with _rules_. For example, assume that centillion is the biggest number in your dictionary. This would not mean that "one centillion one" is not a named number; it's just not a separate word that needs a definition. We haverules for naming numbersthat allow us to use a small list of names and put them together to make names for more numbers than a dictionary could hold. And using those rules, we can make numbers like "a centillion centillion," which is a whole lot larger.

So not only do numbers themselves follow rules that always let us find a bigger one “by induction”, but number names are part of a language system that always lets us *name* a bigger one! But even the usual systems for naming eventually run into trouble.

I took a little digression to talk about the “googol” (no, not Google), because many kids have heard about it:

Then, people can make up words that aren't in dictionaries at all. Back in the 1930's, a man named Edward Kasner, a mathematician, asked his eight-year-old nephew Milton Sirotta to think of a name to give to a large number. Mr. Kasner then said that a "googol" was the number you would write as a one followed by 100 zeroes. Sounds pretty big, doesn't it? But it's not even as large as the centillion I mentioned. However, they then used that new name to give a name to a number that is much, much larger, the "googolplex." This is written as a one followed by a googol of zeroes. Now, a googol is so large that there aren't that many particles in the whole universe; so even if you wrote a zero on every particle in the universe, you couldn't even _write out_ a googolplex, much less count it! So probably "googolplex" is the answer to your second question (though other even larger numbers have been given special names like "Graham's Number"); it is in many dictionaries. But even kids have named larger numbers; what would you think a "googolplexplex" might be?

Bigness gets really out of control, doesn’t it. (That, in fact, was Kastner’s whole point; he wanted to name this huge but easily described number in order to make that point in a book he wrote for kids. The googol is not used by mathematicians apart from this.)

Now, if we can imagine and talk about numbers that are so big no one could ever even write them out, you can see why I said there is nothing that can ever stop numbers from getting larger. Our imaginations are bigger than the whole universe!

Finally, about infinity:

One last comment. The word "infinity" doesn't represent an actual number that is bigger than all the others; "infinite" just means "without end," and is a way of describing something that never comes to an end. There are infinitely many numbers, because there is no last number. And that's really all it means. If you have not already seen it, our FAQ may be of interest: Large Numbers and Infinity http://mathforum.org/dr.math/faq/faq.large.numbers.html

That FAQ includes references about naming large numbers, as well as some details about infinity.

Here’s another fourth grader:

Infinity Questions Is there a bigger number than infinity? What is infinity times infinity? Lee, 10, in Mr. Laurence's Grade 4 class at Simcoe St. School

Dr. Ken answered:

One important thing to keep in mind about infinity is thatit's not really a number, it's something else. It's a concept that means "something bigger than all numbers." Sothere are no numbers bigger than infinity, but that DOESN'T mean that infinity is the biggest number, because it's not a number at all. In fact, there is no biggest number, because if you think you've got it, you can always add one, and you've got something bigger.

At an advanced level, we can talk about larger and smaller infinities, but that requires a different perspective on infinity.

How about multiplying it?

Since infinity isn't really a number, it doesn't always make much sense to multiply it by things, but we can think about doing it anyway. We can say that infinity x infinity = infinity, and keep in mind that we're being a little bit sneaky to even talk about multiplication like that.

This really makes sense only in the context of limits, which we are not considering here.

Kids can ask big questions, and sometimes we can offer big answers; other times, it’s better to keep away from the edge.

]]>I’ll start with a question from 2003 that summarized volume and area formulas for pyramids and cones:

Cones, Pyramids: Surface Area and Volume Formulas How do you get the formula for surface area and volume in the easiest way for cones and pyramids? I know the equations; I just do not understand how to get them.

I first stated the formulas for volume and lateral surface area:

Hi, Aditi. Both formulas are rather easy to remember in this form: V = Bh/3 S = Ps/2

These formulas both apply to pyramid and cone alike, which is why I like them.

In the volume formula, *B* is the area of the base (whether a polygon or a circle), and *h* is the height, the vertical distance from the base to the apex. In the case of the (circular) cone, we can replace *B* with the formula for the area of a circle: $$V = \pi r^2 h$$

In the area formula, *P* is the perimeter of the base (called the circumference in the case of a circle), and *s* is the slant height, the distance from the perimeter to the apex (which is the height of a triangular face in the case of the pyramid).

Then I just referred Aditi to explanations of the derivation of the volume formula, \(V = \frac{1}{3}Bh\):

The volume formula is one of the hardest to actually derive; you can read about that here if you want: Volume of a Cone http://mathforum.org/library/drmath/view/55263.html Volume of a Pyramid http://mathforum.org/library/drmath/view/55041.html But it is easy to understand: the volume is 1/3 the volume of the cylinder or pyramid of the same height with the same base. Since the latter is Bh, the product of the base area and the height, this is simple.

The second link is the one we started with last week for the pyramid; the first does essentially the same thing for the cone, slicing it into cylindrical layers, including a proof of the summation formula we use. Next, I gave a quick derivation of the lateral surface area formula, \(S = \frac{1}{2}Ps\):

The area formula (I've given the lateral area only) for a regular cone or pyramid is easier to prove. Thelateral surface of a pyramid is composed of triangles, all of which have the same height, the slant height s. The height of one of these is bs/2, where b is its base; all together, the bases of the triangles add up to P, the perimeter of the base, so the area is Ps/2. A cone takes the same formula (where the perimeter is called the circumference).

We can think of unwrapping the triangular sides of a pyramid like this:

The (lateral) surface area is the area of *n* triangles with base *b* and height *s*: $$S = n\left(\frac{1}{2}bs\right) = \frac{1}{2}nbs = \frac{1}{2}Ps$$

The cone is similar; we’ll look at it below, but might as well see the picture here for comparison:

Notice the similarity of the formulas: each is something about thebaseas a whole, times ameasurement away from the base,divided by the dimensionof the quantity being calculated. (Volumes are three-dimensional; areas are two-dimensional.) And thevertical heightused for the volume is the distance of the apex from the base whose area it multiplies, while theslant heightused for the surface area is the distance of the apex from the perimeter of the base, whose length it is multiplied by.

That is, the **volume** is found by multiplying the base **area **by the (vertical) distance from the apex to the *center of the base*, divided by 3, while the **surface area** is found by multiplying the base **perimeter **by the (slant) distance from the apex to the *perimeter*, divided by 2.

For a more detailed discussion of the pyramid, we can consider this question, from 1997:

Surface Area of a Pyramid Hello. I would just like to know how to get thesurface area of a pyramid. Thanks.

Doctor Sonya answered:

Hi Daisy! You didn't say what kind of pyramid, but I'm guessing you want the surface area of aregular square pyramid. That's a pyramid with a square base and four sides that are all the same. Can you imagine what such a pyramid looks like? If you have some toothpicks and clay, you can try building one by making four toothpicks into a square (held together at the corners with clay) and then attaching one toothpick to each corner and bringing all four of them together at the top. You can even use mini-marshmallows to hold the toothpicks together if you want. That's how I build my geometry models. Just in case you don't want to build a model,the pyramid has a square base and four equal triangular sides. Do you see why the sides are triangles?

To get the surface area of the pyramid, you need tofind the area of the base and the area of each of the sides, and then add them up.

Note that the formula here will be not just the **lateral (side) surface area** we looked at above, but the **total surface area**, including the base. Different situations can call for either.

Whenever you talk about pyramids, there are two things you have to know: the length of thesidesof the base and theheight. These two things should be given in your problem. The height of a pyramid is how tall it is. If you were to build a hollow pyramid, tie a string to the top, and then let it down into the middle of the pyramid until it hit the floor, the length of the string would be the height. This length is also called analtitude, because it is perpendicular to the floor.

The height *h* and the base side *b* are shown in the picture above.

Now that we know exactly what we are talking about, let's assign some values to our lengths. We'll make them variables because you want a general formula. Call the length ofone side of the base "b"and theheight of the pyramid "h". Thearea of the baseis easy to find. What's the area of a square with side length b?

So far, we have the base area \(B = b^2\).

Now we need to find the areas of the triangles on the sides. Remember that the area of a triangle is: (1/2)(base)(height)The triangles have base b, but what is their height?We can use the Pythagorean Theorem to find the height of the triangles.Picture a triangle inside of the pyramidwith one side straight "down" through the pyramid from the very top to the center of the base, another side from the center of the base to the midpoint of one side of the base, and the third side from the midpoint of one side of the base to the top of the pyramid. Drawing a picture (or using more marshmallows) might help you visualize this. This third side is the height of the triangles we are looking for. Fortunately, we know the other two sides: h and b/2. Think about how we got these. These are also two legs of a right triangle.

Here are our pyramid again, and that triangle inside it, from which we calculate *s*:

Remember that the Pythagorean Theorem says that if the two legs of a right triangle have length A and B, the the length of the hypotenuse, C, can be found with the equation: A^2 + B^2 = C^2. (A^2 means "A squared") The two legs of our right triangle above are of lengths h and b/2, so s^2 = h^2 + (b/2)^2. If we solve this equation for the length of the hypotenuse, we find that it is: SQRT(h^2 + (b/2)^2). So the area of one triangular face is: (1/2)(base)(height) = (1/2)(b)(SQRT(h^2 + (b/2)^2).

So the area of one side of the pyramid is $$A = \frac{1}{2}b\sqrt{h^2 + \left(\frac{b}{2}\right)^2}$$

That means the surface area of the entire pyramid is the areas of the four triangular faces (remember they're all the same) plus the area of the base. You can use this same technique to find surface areas of pyramids with other bases. Let us know if you need some extra help!

Our formula for the total surface area is $$T = b^2 + 4\cdot\frac{1}{2}b\sqrt{h^2 + \left(\frac{b}{2}\right)^2}\\ = b^2 + 2b\sqrt{h^2 + \left(\frac{b}{2}\right)^2}$$

We often express this in two steps: $$T = b^2 + 2bs$$ $$s = \sqrt{h^2 + \left(\frac{b}{2}\right)^2}$$ This, of course, applies specifically to the **square** pyramid. And I find it easier to remember the base area and the lateral area separately, rather than memorizing a total area formula.

Here’s a quicker explanation of the surface area of a square pyramid, which will allow us to find a more general formula:

Surface Area of Pyramids We understand how to find the surface area, but we cannot figure outhow to determine the slant. The information given is the height and the base edge.

Andy already has a formula, probably something like our \(S = \frac{1}{2}Ps\). Here the issue is how to find that slant height *s*, much as we did in the last answer. Not knowing what kind of pyramid Andy had in mind, but guessing that it was not a really hard kind, I again chose to go with a square pyramid: (I’m going to change the variable I used for the base side to *b*, for consistency.)

I'll assume you are dealing with aright square pyramid; for other bases you can work it out almost the same way. Here's our pyramid, with base edge b and height h: + //|\ ///| \ / // | \ / // | \ / / / | \ / / / | \ / / / | \ / s/ / |h \ / / / | \ / / / | \ / / / | \ +-----/---/-------|-----------\ \ / / | \ \ / / | \ +----/---------+ \ \ / r \ \ / \ +-----------------------------+ b The height of the slanted face, s, is the hypotenuse of a triangle with legs r = b/2 and h. You can use the Pythagorean Theorem to find s, s = sqrt(h^2 + (b/2)^2) and then find the area of each face, bs/2.

I used the variable *r* for half the base edge because it is sort of like a radius. It would not be half the edge for other pyramids; in general it is called the **apothem** of the polygon: the distance from the center to the middle of an edge. The important thing is that from that and the height, we can find the slant height.

You may notice that if you want the length of the slanted edge for some reason, you can use the Pythagorean Theorem again. If the base were triangular, you would have to find the apothem r by different means. I'll let you handle that if you need to.

Let’s do the general case. First, the lateral surface area of a pyramid whose base is a regular *n*-gon (*n*-sided polygon) is *n* times \(\frac{1}{2}bs\), which we can simplify to \(\frac{1}{2}nbs = \frac{1}{2}Ps\), since the perimeter *P* is just *n* times the base side *b*. Then, our formula for the slant height in terms of the apothem *r*, becomes \(s = \sqrt{h^2 + r^2}\), so the final formula for the lateral surface area of any regular pyramid is $$S = \frac{1}{2}nb\sqrt{h^2 + r^2} = \frac{1}{2}P\sqrt{h^2 + r^2}$$

Now let’s turn to the cone, with this question from 1999:

Lateral Surface of a Cone What is the equation for the lateral surface of a cone?

I answered with a picture like the one I showed earlier:

I can take your question two ways: theformula for the lateral surface area, or theequation of the surface itself(involving x, y, and z coordinates). I'll assume it's the former. You can find the formula on this page: http://mathforum.org/dr.math/faq/formulas/faq.cone.html The area is pi*r*s, where r is the radius of the base, and s is the length of the side (from the vertex to the base). You can figure it out without too much trouble if you think of the surface as made of a piece of paper rolled up. Unroll it flat, and it will be a sector of a circle: ********* ****** ****** *** / ** / ** / * / * /s * / * / * / * / * +---------------------* * * * * * * * * * * * * ** ** ** ** *** *** 2 pi r ****** ****** *********

Here as before, *s* is the slant height; and the circumference of this sector is the circumference of the base, whose radius is *r*. Don’t confuse these two different radii!

The arc length of the whole circle would be 2 pi s; but you only have as much as will fit around the base of the cone, 2 pi r. That means you have 2 pi r r ------ = --- 2 pi s s of the whole circle, so the area is r --- * pi s^2 = pi * r * s s

The angle at the center of the sector is a fraction *r*/*s* of a complete circle (360 degrees), so the area is that fraction of the area of a circle.

Compare this \(S = \pi r s\) to the area of the pyramid, \(S = \frac{1}{2}Ps\). If you replace the perimeter P with the circumference \(C = 2\pi r\), do you see the similarity? Our final question will make that explicit in providing an alternative path to the formula.

For an alternate approach, obtaining the formula for a cone from that for a pyramid, consider this question from 2003:

Surface Area of Cones and Pyramids Can the method for finding the surface area of apyramidbe used as well to find the surface area of acone?

Doctor Jerry answered:

Hi Nadiya, Yes, if you're willing to be a bit loose with procedures. The area of a pyramid is 2*b*s, where b is the side of the base and s the slant height. I suppose one can find this by looking at one triangle, whose area is (1/2) * height * base = (1/2)*s*(b) There are four of these, for a total area of 2*b*s.

Doctor Jerry is assuming a square pyramid (*n* = 4). In general, this would be our \(S = \frac{1}{2}Ps = \frac{1}{2}nbs\).

Now we apply this to a cone, thinking of its circular base as a polygon with a very large number of sides:

On the cone, divide the circle into a large number of equal pieces, say n of them. We know that each has a base of length 2*pi*r/n. So, this is a. The slant height is s. Each base can be connected to the vertex, making a "triangle." The resulting area is (1/2) * s * 2*pi*r/n There are n of these for a total area of pi*r*s. I believe that Kepler used these kinds of (risky) arguments.

Putting it more succinctly, the *P* in \(\frac{1}{2}Ps\) is replaced by \(C = 2\pi r\), and we get \(S = \frac{1}{2}2\pi rs = \pi r s\). The fact that n disappeared from the final formula tells us that it applies even when *n* “becomes infinite”.

Many of our explanations in this series have been this sort of “risky” near-proofs, which would need extra work to provide full proofs, but are good enough for basic understanding!

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