Chapter 9

Right-Triangle Trigonometry

Right Makes Might: Solving Right Triangles

In the next few chapters you’ll learn how to “solve” any determined triangle – to find, that is, its missing parts. This is trigonometry’s basic problem, its historical raison d’être. In this chapter, you’ll specifically learn how to solve right triangles, which are the polygonal world’s atoms: any polygon can be dissected into triangles, and any triangle can be dissected into right triangles.

The crucial fact about right triangles, the fact that provides the basis for all of trigonometry, is this: A right triangle’s proportions are entirely determined by just one of its acute angles. For example, all right triangles with a $ 30^{} $ angle are similar (by AA), so they all have the same proportions. $ ^{*} $

Consider a related phenomenon. All circles are similar, so all circles have the same proportions. In particular, the Circumference-to-Diameter ratio is the same in all circles. This ratio’s value is so important that we give it a special symbol: $ $ . As the figure at right shows, $ $ is a bit greater than 3. The number $ $ , of course, is an old friend. Let us now meet some of our friend’s friends, defined similarly (pun intended), based not on circles, but on right triangles.

Let $ $ be any fixed acute angle. As discussed above, all right triangles containing $ $ are similar, and thus have the same proportions. In particular, the ratio of the leg opposite $ $ to the hypotenuse is the same in all such triangles. This ratio is called “the sine of theta,” which we write symbolically as $ $ .

Definition. For each acute angle $ $ , we define $ $ as follows: In any right triangle containing $ $ ,

\[ \mathbf{s i n}\theta=\frac{\mathrm{o p p o s i t e}}{\mathrm{h y p o t e n u s e}}, \]

where “opposite” refers to the leg lying opposite $ $ .

For example, the figure at right suggests that in any right triangle with a $ 22^{} $ angle, the leg opposite that angle goes into the hypotenuse about 2.7 times. Thus, $ ^{} /2.7 $ . Of course, this is crude approximation based entirely on eyeballing a figure, but it should clarify the simple idea behind sine.

The ancient Greek astronomers were the first to construct sine tables. For instance, Ptolemy managed, around 150 A.D., to find values of $ ^{} $ , $ ^{} $ , $ ^{} $ , $ ^{} $ , and so forth (up to $ ^{} $ ), accurate to five decimal places. His table – and, better still, his explanation of how he constructed it – survives in his masterpiece, the Almagest, which was the Bible of geocentric astronomy. Over the centuries, others around the world produced increasingly refined tables that one could use, say, to approximate $ ^{} $ to ten decimal places. Until the 1980s, anyone studying trigonometry would have done so with a table of sines at hand. Inexpensive handheld scientific calculators have since displaced the tables, but they serve exactly the same purpose.

Most scientific calculators will approximate the sine of any angle to 9 decimal places or so. Punch sin $ 22^{} $ into yours, for example, and it will tell you that $ ^{} $ . Later (at the end of Chapter 12), I’ll describe in broad terms what the calculator does to produce that approximation. For now, however, let us pause to appreciate what Ptolemy’s table of sines (or its modern descendant, the scientific calculator) accomplishes: It allows us to solve any determined right triangle.

The idea is simple: $ $ relates three parts of a right triangle: an acute angle $ $ , its opposite leg, and the hypotenuse. If we know two of these three parts, a table (or calculator) can give us the third.

Example 1 (Using a calculator to find a side). Solve the triangle at right.

Solution. The remaining angle is obviously $ 68^{} $ .

If we call the hypotenuse c, then sin 22° = 5/c.

Equivalently, $ c = 5/^{} $

Running this through a calculator (or a table of sines) yields $ c $ .

Now that we have two sides, we can get the third side by using the Pythagorean Theorem. Doing so, we find that the long leg is approximately 12.38 units long.

Example 2. (Using a calculator to find an angle). Solve the triangle at right:

Solution. The hypotenuse is $ $ by the Pythagorean Theorem.

Call the larger acute angle θ. Then sin θ = 5/√29 ≈ 0.9285.

If we were using a table of sines, we’d scan it until we found the angle whose sine is 0.9285; this angle would be our approximation for $ $ . To accomplish the same end with a calculator, we use its “inverse sine” button, which looks something like this: $ $ . In the context of solving right triangles, the inverse sine of x simply means “the angle whose sine is x.” Thus, since we seek the angle whose sine is $ 5/ $ , we type $ ^{-1}(5/) $ into a calculator, which tells us that

\[ \theta=\sin^{-1}\left(5/\sqrt{29}\right)\approx\mathbf{68.2^{\circ}}. \]

The remaining angle is therefore approximately $ 90^{}-68.2{}=21.8^{} $ .

Exercises

1. Use a calculator to solve the three right triangles at right, finding their angles exactly, and the lengths of their missing sides to the nearest hundredth of a unit.

2. Use a calculator to solve the three right triangles at right, finding their side lengths exactly, and their angles to the nearest tenth of a degree.

3. Make up your own right triangle with enough information to determine it, and solve it.

Measuring Inaccessible Distances

You’ve already learned enough to grasp one of trigonometry’s basic applications: measuring inaccessible distances. The idea is to identify a determined triangle, one side of which is the distance you want. Then use trigonometry to find this distance.

Example (The obligatory flagpole problem). You want to measure a flagpole but don’t want to

shimmy up to the top, trailing a tape measure behind you. How can you do it?

Solution. Walk ten feet from the pole, then measure the angle of inclination from where you stand to the pole’s top. Let’s say that this turns out to be $ 63^{} $ . We now have a determined right triangle (by ASA), so we can solve it with trigonometry.

The triangle’s other acute angle is $ 27^{} $ . If we let c be the hypotenuse, then

$ ^{} = 10/c $ . Equivalently, $ c = 10/^{} $ feet.

We now have two sides of the triangle, so we can get the third with the Pythagorean Theorem. Doing so, we find that the flagpole’s height is about 19.626 feet, which works out to about 19’7.5”, as you should verify.

After you’ve learned some further trigonometric techniques, we’ll forget the flagpole and take on much more impressive applications: measuring inaccessible astronomical distances. Even then, however, the spirit of the trigonometric game will remain the same – finding and solving determined triangles.

Exercises

4. Had we walked only 5 feet from the pole, what would the angle of inclination have been?

[Hint: Don’t forget, you’ve already found the pole’s height.]

5. By measuring the lengths of the shadows cast by a flagpole and a yardstick (the yardstick, like the flagpole, should be perpendicular to the ground), one can find the flagpole’s height without trigonometry. Explain how.

6. A twist on the flagpole problem (as solved in exercise 5) appears in the Sherlock Holmes story “The Adventure of the Musgrave Ritual.” Read it if you feel so inclined.

7. Moses needs to measure the distance across the Jordan river, but cannot cross it. He stares at a tree directly across from him on the opposite bank. “Oy gevalt,” he mutters. “Would that I were back by the fleshpots of Egypt.” After walking twenty feet along the riverbank, he turns and looks back at the tree. Suddenly inspired, he measures the angle shown in the figure, which turns out to be $ 53^{} $ . To the nearest inch, how wide is the river?

I nterlude: Three Exact Values of Sine

Thus far, we’ve settled for approximate values of sine from a calculator. There are, however, certain angles whose sines we can compute exactly. This is the sort of thing that warms the cockles of a mathematician’s heart. In this section we’ll consider three such angles, all of which appear in simple shapes: half of a square and half of an equilateral triangle.

Let’s start by splitting an equilateral triangle in half as in the figure at right. The result is a “30-60-90 triangle”. All such triangles are similar (by AAA), so we’re free to investigate a 30-60-90 triangle of any size. I’ll make the short leg 1 unit. (You could make it any length you please, but this choice will simplify our work.) The other sides’ lengths are now determined. Clearly, each side of the equilateral triangle is equal to two copies of the right triangle’s short leg (see the figure). Thus, the right triangle’s hypotenuse, being one of the equilateral triangle’s sides, is 2 units. The Pythagorean Theorem then yields the long leg’s length: $ $ units.

From the figure above, we can read off two exact values of sine:

\[ \sin30^{\circ}=\frac{1}{2},\qquad\sin60^{\circ}=\frac{\sqrt{3}}{2}. \]

You should commit these to memory, and if you ever forget them, you should be able to re-derive them in a matter of seconds by thinking through the preceding argument.

Finding sin $ 45^{} $ is even easier, because we can find this angle in half of a square – in a right isosceles triangle, as in the figure. Letting each leg of such a triangle be 1 unit, the Pythagorean Theorem tells us that the hypotenuse must be, $ $ . Hence,

\[ \mathbf{s}\mathbf{i}\mathbf{n}4\mathbf{5}^{\circ}=\frac{\mathbf{1}}{\sqrt{2}}\left(=\frac{\sqrt{2}}{\mathbf{2}}\right). \]

This too you should memorize – and understand.

Exercise

8. Finding an exact expression for a value of sine (i.e. in terms of radicals) is difficult, and in most cases, impossible. It is therefore cause for celebration when we can do it. In this problem, you’ll find two more exact values (values you need not memorize!) with an ingenious geometric argument.

Draw a square and an equilateral triangle on the same base. Through the top vertex of the triangle, draw a line perpendicular to the base. Now, given the labels I’ve provided in the figure…

1. Find the angles in $ AED $ and $ DEF $ . [Hint: $ AED $ is a special type of triangle.]

2. Let AG = 1. Find $ DEF’ $ ’s sides, which will yield the sines of its acute angles.

SOH CAH TOA: The God of Right Triangles

All right triangles containing a given acute angle $ $ have the same proportions. The various ratios of their sides have special names, one of which you already know. Here are three we’ll use over and over again.

Definition. For each acute angle \(\theta\), we define the following trigonometric ratios:

\[ \sin\theta=\frac{opposite}{hypotenuse},\qquad\cos\theta=\frac{adjacent}{hypotenuse},\qquad\tan\theta=\frac{opposite}{adjacent}, \]

where the terms in each ratio refer to the sides of any right triangle containing $ $ . (“Opposite” and “adjacent” refer to the positions of the triangle’s legs relative to $ $ .)

Most people remember these three ratios (which are read “sine of \(\theta\)”, “cosine of \(\theta\)”, and “tangent of \(\theta\)”) by invoking the awesome name of the god of right triangles, SOH CAH TOA.*

We can solve any determined right triangle using sine alone, but having cosine and tangent at our disposal considerably shortens our work. Consider our flagpole example from two pages ago. To find its height using sine alone, we first had to shift our attention to the acute angle that we weren’t given; we then used sine to find the hypotenuse, and the Pythagorean Theorem to get the flagpole’s height. Compare the following much simpler solution.

Example. Find the height of the flagpole at right.

Solution. We have the leg Adjacent the $ 63^{} $ angle. We want the Opposite leg. Now we chant SOH CAH TOA’s name… and we remember: A-ha! We should work with tangent, because Tangent relates the Opposite and Adjacent legs. In particular, if we let x be the flagpole’s height, then

\[ \tan63^{\circ}=x/10. \]

Or equivalently,

\[ x=10\tan63^{\circ}\approx19.6261\ feet. \]

That is, the flagpole is about 19 feet, 8 inches tall. Praise be to SOH CAH TOA!

Exercises

9. Find – without a calculator, obviously – exact values for the cosine and tangent of $ 30^{} $ , $ 60^{} $ , and $ 45^{} $ .

10. Find the missing sides in each of the triangles at right without using the Pythagorean Theorem.

11. A right triangle has sides 3, 4, and 5. Use the inverse tangent to approximate its smallest angle to the nearest tenth of a degree. (The inverse tangent is analogous, of course, to the inverse sine.)

I dentity Crisis – and Why Cosine is Called Cosine

The two acute angles in any right triangle are, of course, complements. Realizing this, a glance at the figure reveals that the cosine of any angle $ $ is the sine of its complement.

(Because in terms of the labels on the figure at right, they both equal a/c.)

In fact, the “co” in cosine stands for complement: cosine literally means “sine of the complement”. We’ve just established the first trigonometric identity you’ll see in this course: for any acute angle $ $ ,

\[ \mathbf{\cos\theta}=\mathbf{\sin(90^{\circ}-\theta)}. \]

We shall use trigonometric identities much in the same way that we use algebraic identities such as $ a^{2}-b{2}=(a-b)(a+b) $ . Each identity gives us two forms in which we can express the same thing. Sometimes, one form is more convenient than the other. We use whichever form best suits the occasion.

Staring hard at our generic right triangle in the little figure above, we discern another trigonometric identity: For each acute angle $ $ ,

\[ \mathbf{t a n}\theta=\frac{\sin\theta}{\cos\theta}. \]

To prove this, just call on SAH CAH TOA to re-express each side of this equation in terms of a, b, and c: As you should verify, each side equals b/a. Hence they are equal to one another as claimed.

Learn these two identities well. We’ll use them frequently.

Exercises

12. Rewrite each of the following cosines as sines. (For example, $ ^{} = ^{} $ .)

1. $ ^{} $

2. $ ^{} $

3. $ ^{} $

13. By now, you should know the sines of the special angles $ 30^{} $ , $ 60^{} $ , and $ 45^{} $ by heart. You should know these angles’ cosines, too. The good news is that you don’t have to memorize these separately. Explain why.

14. Prove that $ =(90^{}-) $ for all acute angles $ $ .

[Hint: Using the figure above, rewrite the trigonometric expressions in this equation in terms of a, b, and c.]

15. Rewrite each of the following sines as cosines: a) $ ^{} $ b) $ ^{} $ c) $ (+ 20^{}) $

16. Rewrite each of the following tangents in terms of sines and cosines: a) tan 10° b) tan 25° c) tan 88°

17. You should now be able to produce the exact tangents of $ 30^{} $ , $ 60^{} $ , and $ 45^{} $ in two different ways. Do so.

[Hint: One way is to draw the appropriate triangles. The other is to use the identity for tangent you just learned.]

These three values are worth memorizing.

18. Simplify the following expressions.

1. $ $ b) $ $ c) $ $

19. Prove that $ ()^{2}+(){2}=1 $ for all acute angles $ $ . [Hint: See the hint for Exercise 14.]

20. What does $ $ have to do with Sherlock Holmes?

The Reciprocal Trig Functions

The reciprocals of the three basic trigonometric ratios have special names and symbols:

Definitions. For each acute angle $ $ ,

The reciprocal of $ $ is called the secant of $ $ .

The reciprocal of $ $ is called the cosecant of $ $ .

The reciprocal of $ $ is called the cotangent of $ $ .

(sec $ = 1/$ )

(csc $ = 1/$ )

(cot $ = 1/$ )

Thus, for example, $ ^{} = = = ( = ) $ .

The reciprocal functions are strictly heterosexual: sec goes with cos, and csc goes with sin. This completes our collection of trigonometric functions.

Exercises

21. Find exact expressions for the secant, cosecant, and cotangent of $ 30^{} $ , $ 60^{} $ , and $ 45^{} $ .

22. Is $ $ a ratio of sides in a right triangle containing $ $ ? If so, which ratio of sides? If not, why not?

23. Sine and cosine are called cofunctions because, as you’ve learned, the sine of any acute angle is the cosine of its complement, and vice-versa. Secant and cosecant, as their names suggest, make up a second pair of cofunctions, while tangent and cotangent make up a third pair of cofunctions.

Your problem: Prove that secant and cosecant truly are cofunctions. Then do it for tangent and cotangent.

[Hint: Adapt the strategy we used on the previous page to prove that sine and cosine are cofunctions.]

24. Use cofunctions to rewrite the following in terms of angles less than $ 45^{} $ (e.g. $ ^{} = ^{} $ ).

1. $ ^{} $ b) $ ^{} $ c) $ ^{} $ d) $ ^{} $ e) $ ^{} $ f) $ ^{} $

25. Use cofunctions to rewrite the following in terms of angles greater than $ 45^{} $ .

1. $ ^{} $ b) $ ^{} $ c) $ ^{} $ d) $ ^{} $ e) $ ^{} $ f) $ ^{} $

26. True or False? (Explain your answers.) For every acute angle ,

1. $ = (90^{} - ) $ b) $ = (90^{} - ) $ c) $ = 1/(90^{} - ) $ d) $ = 1/$ e) $ = 1/$ f) $ = 1/$ g) $ /= $ h) $ = $ i) $ /= $

27. Simplify the following expressions.

1. $ $ b) $ $ c) $ $ d) $ - $

28. Cosine is just “sine with a twist”, since $ =(90^{}-) $ . The same is true of tangent (with a bigger twist), since $ =/[(90^{}-)] $ . Show how each of the reciprocal trig functions can be understood as twisted versions of sine, too.

[Moral: Trigonometry is, in essence, the study of the sine function.]

Basic Right-Angle Trig: Shortcuts

What Romantic terminology called genius or talent or inspiration or intuition

is nothing other than finding the right road empirically, following one’s nose,

taking shortcuts…

• Italo Calvino, “Cybernetics and Ghosts”, from The Uses of Literature.

In the previous chapter, you learned two shortcuts for applying the Pythagorean Theorem in your head. There are analogous shortcuts for solving right triangles. This is a very good thing, because solving right triangles by appealing to SOH CAH TOA involves a fair amount of drudgery: First you must introduce a new symbol for the unknown that you seek; then you must set up an equation; finally, you must solve that equation. Once you’ve gone through that mechanical process a few times, it becomes tiresome. Better to condense the process and leave drudgery to the drudges.

The following shortcuts will make your work neater, since they obviate the need to introduce new symbols at every turn. And by condensing the drudgery, they’ll also free your mind to concentrate on the more conceptual parts of whatever trigonometric problem you may be working on.

Shortcut 1. Given an acute angle and the hypotenuse,

The opposite leg is the hypotenuse times the angle’s sine;

The adjacent leg is the hypotenuse times the angle’s cosine.

Proof. Let h and $ $ be the hypotenuse and given acute angle.

By definition, $ = /h $ . Multiplying both sides by h gives us the first result.

By definition, $ = $ . Multiplying both sides by h gives us the second result.

Shortcut 2. Given an acute angle and its adjacent leg,

The opposite leg is the given leg times the angle’s tangent;

The hypotenuse is the given leg times the angle’s secant.

Proof. Let a and $ $ be the hypotenuse and given acute angle.

By definition, $ = /a $ . Multiplying both sides by a gives us the first result.

By definition, $ = $ . Multiplying both sides by a gives us the second result.

These shortcuts are best remembered by their accompanying figures. Bear in mind that they are just SOH CAH TOA in fancy garb. For that reason, whenever I use one, I’ll indicate as much with a stock phrase such as “by basic right-angle trigonometry” or “by basic trig”. You should do as much.

To appreciate these shortcuts, go back and reread our first solution of the flagpole problem, in which we used only sine. Then reread our second solution, in which we called on SOH CAH TOA’s aid. Now consider this third solution, for which we’ll use our $ 2^{nd} $ shortcut.

Example 1. Find the height of the flagpole at right.

Solution. By basic right-angle trig, its height is $ 10 ^{} $ feet.

(Or, rounded to the nearest inch, 19’8”.)

The following problem would be tedious without our shortcuts, but with them, it takes just seconds.

2

Example 2. In the figure at right, find length ED.

Solution.

First, $ BC = 3 ^{} $

Next, $ CD = BC ^{} $ (basic trig on $ ABC $ ).

(basic trig on $ BCD $ ). $ ^{*} $

Finally, $ ED = CD ^{} $ (basic trig on $ ECD $ ). $ ^{} $

In that solution, observe that whenever I applied basic trigonometry, I mentioned the specific triangle to which I applied it. You should do this too whenever you solve a problem involving multiple triangles.

You never have to use these two shortcuts, of course. You could always just call on SOH CAH TOA and work out the ratios. But still, anyone taking calculus or physics should be very comfortable using at least the first of the two shortcuts. Indeed, you’ll probably get funny looks in future classes you don’t know it. In physics, for example, we use that first shortcut all the time to resolve vectors into their horizontal and vertical components. If you don’t know that first shortcut, and instead must pray to SOH CAH TOA whenever such a triangle arises, your physics teacher is going to look at you askance – not because you’re doing anything wrong, but because you should be able to do that quickly by now. The second shortcut can also be useful, and it’s a nice one with which to impress your friends, but realistically, no one will look at you askance if you don’t know it. But still, it’s so simple, so why not take ten minutes to learn it and get comfortable using it?

Exercises

29. Solve each of the triangles at right, using the shortcut formulas introduced in this section to find the missing sides.

[Hint for the first one:

Get the remaining acute angle first.]

30. In the figure below, solve $ AEF $ .

31. In the figure below, solve $ AEF $ .

32. Explain why the shortcuts introduced in this section work.

Eratosthenes Measures the Earth

The overriding sensation I got looking at the earth was, my god that little thing is so fragile out there.

• Michael Collins, astronaut.

Eratosthenes, a $ 3^{rd} $ -century B.C. Alexandrian geographer, mathematician, and jack of all intellectual trades, knew that at noon on a certain day, the sun would be directly overhead in the city of Syene, 500 miles due south of Alexandria. Because Alexandria and Syene lie on the same meridian, noon occurs simultaneously at both cities. At noon on the appointed day, Eratosthenes propped his staff up perpendicular to the ground and measured the angle of elevation from the tip of its shadow to the top of his staff. With this angle (82.8°), he determined Earth’s circumference.

To understand how he did it, consider the figure at right, which shows the meridian of Earth containing Alexandria and Syene (A and S), Eratosthenes’ staff (AB: not to scale!), and several solar rays, which are effectively parallel since the Sun and the Earth are both so tiny compared to the vast distance between them. When the sun is directly overhead at Syene, Eratosthenes’ measurement of $ AB $ lets him simultaneously measure an angle at the center of the Earth. Behold:

\[ A\hat{E}S=C\hat{B}A=90^{\circ}-A\hat{C}B=7.2^{\circ}. \]

This is certainly the cleverest application of zigzag angles you’ll ever see. Because this central angle of $ 7.2^{} $ is just 1/50 of a full $ 360^{} $ , it follows that the circular arc that it subtends, arc AS, must be 1/50 of the Earth’s full circumference. Hence, Earth’s circumference must be approximately

\[ 50(500)=25,000\text{miles}. \]

Dividing this by $ 2$ reveals the Earth’s radius: about 4000 miles, a figure that we’ll use on the next page, when we apply trigonometry to astronomy.

Exercise

33. Every circle has a unique tangent at each of its points. The circle’s perfect symmetry guarantees that each tangent is perpendicular to the radius it touches. We can use this important property to smuggle right angles – and hence trigonometry – into problems about circles. We’ll do this several times in the next few pages.

Prove that if we draw two tangents to the same circle from a point outside the circle, then the distances from that point to the two points of tangency must be equal. [Hint: Find two congruent right triangles.]

How High the Moon

Deedly ap a do’dn ah b’deeba dee ah…

• Ella Fitzgerald

The methods I will use to compute the moon’s size and distance from Earth involve practical difficulties that I shall cheerfully ignore; I shall also play fast and loose with significant figures. Such scientific details do not interest me here. I wish instead to demonstrate a small miracle: Trigonometry allows us, in principle, to calculate these lunar distances using measurements made entirely on the Earth’s surface.

Problem 1. Using only measurements made on Earth, find the distance to the moon.

Solution. First, locate two points on Earth with this property: When someone standing at A sees the moon at the zenith, someone at B would see it on the horizon.

Second, find the distance between these points, measured along the Earth’s surface.

Let us suppose you have done this, either by

taking a long journey between A and B and taking measurements as you go, or by consulting a friendly geographer, who already knows the distance. This will turn out to be about 6186 miles.

Third, find the angle $ AB $ that arc AB subtends at Earth’s center. We need not journey to the center of the Earth to do this. A proportion will suffice:

\[ \frac{A\hat{E}B}{360^{\circ}}=\frac{\mathrm{arc}\;AB}{\mathrm{Earth}^{\prime}\mathrm{s\;circumference}}. \]

Thanks to Eratosthenes, we already know Earth’s circumference: about 25,000 miles. Moreover, we’ve already ascertained that arc AB is about 6186 miles. Substituting these values into the proportion, we can solve for $ AB $ . (It is a bit larger than $ 89^{} $ , as you should verify.)

Fourth and last, note that $ EBM $ is now determined by ASA. By basic trigonometry, we have $ EM = EB (AB) $ . We know $ AB $ , and we know that EB is about 4000 miles (Eratosthenes again), so we can compute EM, the distance to the moon, as required. If you carry out the computations, you’ll see that the moon is nearly a quarter million miles from Earth.

Problem 2. Using only measurements made on Earth, find the moon’s radius.

Solution. From a point E on Earth, measure the angle that the full moon subtends at your eye. (This is approximately $ 0.5^{} $ .)

Since the distance to the moon is about 250,000 miles, right triangle $ EMT $ is determined by ASA. Applying basic right-angle trigonometry, we obtain the moon’s radius:

$ TM ,000 (0.25^{}) $ , which is about 1000 miles.

Exercises

34. We’ve now seen that the Earth’s radius is about four times that of the moon. What does this imply about the relative volumes of the Earth and moon?

35. A body diagonal of a cube is a line segment running through the cube’s center and joining two opposite corners. In contrast, a face diagonal joins opposite corners of one of the cube’s faces. (In the figure at right, and $ AC’ $ is a body diagonal, AC is a face diagonal.) Find the angle between a body diagonal and a face diagonal.

36. In the figure at right, ten small circles (of the same size) are packed into one large circle; each small circle touches the large one at one point, and adjacent small circles touch at a single point as well. If the large circle’s radius is 1, find the radii (to 3 decimal places) of the small circles.

[Hints: First, a fact from geometry: When circles touch each other at a single point, the line through their centers passes through their point of contact. Second, a problem-solving technique: Find two expressions for the same thing (one of which involves the unknown you seek), then equate them and solve for the unknown.]

37. When you gaze out at the horizon, you are not, strictly speaking, looking at the horizon, but rather at your horizon. In particular, if a 6-foot-tall man and his 5-foot-tall wife stand side by side at the seashore and stare out at the line where sky and sea meet, they are actually gazing at slightly different horizons.

1. Explain why this is so.

2. Assuming that Earth’s radius is 4000 miles, how distant is the man’s horizon, as measured by the straight line joining his eye to the horizon? How distant is his wife’s horizon? (Both answers should be given in miles and feet. 1 mile = 5280 feet.)

3. How high must one be to see 5 miles to the horizon? To see 10 miles to the horizon?

4. Relative to the man in part (b), which distance is greater: the distance of his eyes to the horizon, measured by his line of sight, or the distance of his feet to the horizon, measured along the earth’s curved surface? How much greater is the greater one? Assume that Earth is perfectly spherical.

[Hint: When we found the distance to the moon and the Earth’s size, we used a proportion to relate arcs of a circle to the angles they subtend at the circle’s center. This technique will work here.]

5. Read Chapter XXXV, “The Mast-Head,” of Moby Dick. No – Read the whole book, ye landlubber!

38. Thirty feet separate Quixote and Sancho, who stand on one bank of a river and contemplate a Spanish gentleman with a maimed hand sitting quietly on the opposite bank. After some time, they turn towards one another and exchange a knowing look. If Quixote turns $ 65^{} $ to shift his gaze from the maimed Spaniard to Sancho, and Sancho turns $ 34^{} $ to meet his master’s gaze, how wide is the river?

[Hint: Recall the second hint to Exercise 36.]

39. Find exact values for the sine, cosine, and tangent of $ 72^{} $ and $ 18^{} $ .

[Hint: Ponder the pentagram. See exercise 17 in the previous chapter. Can you find a $ 72^{} $ angle there?]

40. Eratosthenes found the Earth’s radius using only basic geometry. Here’s a trigonometric method that accomplishes the same thing. A man stands on a mountaintop 3 miles above sea level and carefully measures the angle of depression to the ocean horizon ( $ ZH $ in the figure), which he finds to be $ 2.23^$ .

From this, compute the Earth’s radius.

41. By definition, $ $ is the number of times that any circle’s diameter fits into its circumference. A reasonably careful drawing reveals that, as a first crude approximation, $ $ .

1. Read 1 Kings 7:23–26, which describes a curious vessel in Solomon’s Temple. How does $ $ relate to its dimensions? Much scholarly ink has been spilled over this apparently innocuous passage.

2. Everyone “knows” that $ $ ’s decimal expansion never ends, that it follows no discernible pattern, and that it begins 3.14159265358979. $ ^{*} $ Of course, most of us know these facts about $ $ only secondhand – the way we “know” most things. Consider how it is that you know each of the following:

\[ iii.\sin60^{\circ}=\frac{\sqrt{3}}{2}.\qquad iv.\pi\approx3.14159... \]

Assuming you are truly learning trigonometry (and not just memorizing a few facts and techniques), your answer to (iii) should differ significantly from your answer to (iv). Does it?

3. So where do those fabled digits of $ $ (as of 2023, over 100 trillion digits have been computed) come from? Most methods for generating them depend on infinite series, which is a calculus topic, but trigonometry will take you half way there. Consider a regular polygon inscribed in the unit circle (i.e. a circle with radius 1). The polygon’s area is approximately (but slightly less than) the circle’s area, which, in turn, is $ $ . Thus, any decimal approximation for the polygon’s area will give us a lower bound for the decimal expansion of $ $ . The more sides in the polygon, the better the approximation of $ $ . Explain why this is so. Then use trigonometry to find the area of a regular 180-sided polygon inscribed in the unit circle.

4. To obtain an upper bound for $ $ , we can consider a regular polygon circumscribed around the unit circle. Use trigonometry to find the area of a circumscribed regular 180-sided polygon, and explain how this, combined with your work in part (c) shows that $ $ ’s decimal expansion must begin 3.14.

5. Your numerical work in parts (c) and (d) required a calculator – but how does your calculator “know” how to evaluate trigonometric functions? You should now have some insight into why $ $ , but… you should also recognize that we’ve just relocated the mystery: we’ve reduced the problem of approximating $ $ to the problem of approximating the values of trig functions. Later, I’ll explain what your calculator is doing… which will, naturally, relocate the mystery again! Until then, ponder the problem.