# Right Triangles

Let’s agree again to the standard convention for labelling the parts of a right triangle. Let the right angle be labelled C and the hypotenuse c. Let A and B denote the other two angles, and a and b the sides opposite them, respectively.

#### Solving right triangles

We can use the Pythagorean theorem and properties of sines, cosines, and tangents to solve the triangle, that is, to find unknown parts in terms of known parts.
• Pythagorean theorem:   a2 + b2 = c2.
• Sines:   sin A = a/c, sin B = b/c.
• Cosines:   cos A = b/c, cos B = a/c.
• Tangents:   tan A = a/b, tan B = b/a.
Let’s first look at some cases where we don’t know all the sides. Suppose we don’t know the hypotenuse but we do know the other two sides. The Pythagorean theorem will give us the hypotenuse. For instance, if a = 10 and b = 24, then c2 = a2 + b2 = 102 + 242 = 100 + 576 = 676. The square root of 676 is 26, so c = 26. (It’s nice to give examples where the square roots come out whole numbers; in life they usually don’t.)

Now suppose we know the hypotenuse and one side, but have to find the other. For example, if b = 119 and c = 169, then a2 = c2 – b2 = 1692 – 1192 = 28561 – 14161 = 14400, and the square root of 14400 is 120, so a = 120.

We might only know one side but we also know an angle. For example, if the side a = 15 and the angle A = 41°, we can use a sine and a tangent to find the hypotenuse and the other side. Since sin A = a/c, we know c = a/sin A = 15/sin 41. Using a calculator, this is 15/0.6561 = 22.864. Also, tan A = a/b, so b = a/tan A = 15/tan 41 = 15/0.8693 = 17.256. Whether you use a sine, cosine, or tangent depends on which side and angle you know.

#### Inverse trig functions: arcsine, arccosine, and arctangent

Now let’s look at the problem of finding angles if you know the sides. Again, you use the trig functions, but in reverse. Here’s an example. Suppose a = 12.3 and b = 50.1. Then tan A = a/b = 12.3/50.1 = 0.2455. Back when people used tables of trig functions, they would just look up in the tangent table to see what angle had a tangent of 0.2455. On a calculator, we use the inverse trig functions named arctangent, arcsine, and arccosine. Usually there’s a button on the calculator labelled “inv” or “arc” that you press before pressing the appropriate trig button. The arctangent of 0.2455 is 13.79, so the angle A is 13.79°. (If you like, you can convert the 0.79 degrees to minutes and seconds.)

That’s all there is to it.

#### The other three trigonometric functions: cotangent, secant, and cosecant

For most purposes the three trig functions sine, cosine, and tangent are enough. There are, however, cases when some others are needed. In calculus, secant is frequently used. You might ask, “why six trig functions?” It’s a kind of symmetry. There are six ways of making ratios of two sides of a right triangle, and that gives the six functions:
1. sin A = a/c (opp/hyp)
2. cos A = b/c (adj/hyp)
3. tan A = a/b (opp/adj)
4. cot A = b/a (adj/opp)
5. sec A = c/b (hyp/adj)
6. csc A = c/a (hyp/opp)
You can see by the listing that cotangent (abbreviated cot, or sometimes ctn) is the reciprocal of tangent, secant (abbreviated sec) is the reciprocal of cosine, and cosecant (abbreviated csc, or sometimes cosec) is the reciprocal of sine. They’re pretty much redundant, but it’s worthwhile to know what they are in case you come across them. Note that cotangents are tangents of complementary angles, which means that cot A = tan B, and cosecants are secants of complementary angles, and that means that csc A = sec B.

These other three functions can also be interpreted with the unit circle diagram.

We’re considering the angle AOB. Recall that its tangent is the line AC. By symmetry the tangent of the angle FOB is the line FG, but FOB is the complementary angle of AOB, hence, the cotangent of AOB is FG.

Next, to interpret secants geometrically. The angle AOB appears in the triangle COA as angle AOC, so sec AOB = sec AOC = hyp/adj = OC/OA = OC. There you have it–the secant is the line from the center of the circle to the tangent line AC. The reason it is called the secant is because it cuts the circle, and the word “secant” comes from the Latin word meaning “cutting.”

Similarly, the cosecant of the angle AOB is the line OG from the center of the circle to the cotangent line FG. #### Exercises

Note: as usual, in all exercises on right triangles, c stands for the hypotenuse, a and b for the perpendicular sides, and A and B for the angles opposite to a and b respectively.

26. In each of the following right triangles of which two sides are given, compute the sin, cos, and tan of the angles A and B. Express the results as common fractions.
(i). c = 41, a = 9.
(ii). c = 37, a = 35.
(iii). a = 24, b = 7.

31. In a right triangle c = 6 feet 3 inches and tan B = 1.2. Find a and b.

34. a = 1.2, b = 2.3. Find A and c.

42. a = 10.11, b = 5.14. Find B and c.

In the next few problems, the triangles aren’t right triangles, but you can solve them using what you know about right triangles.

61. In an oblique triangle ABC, A = 30°, B = 45°, and the perpendicular from C to AB is 12 inches long. Find the length of AB.

67. If the side of an equilateral triangle is a, find the altitude, and the radii of the circumscribed and inscribed circles.

202. From the top of a building 50 feet high the angles of elevation and depression of the top and bottom of another building are 19° 41' and 26° 34', respectively. What are the height and distance of the second building.

207. From the top of a lighthouse 175 feet high the angles of depression of the top and bottom of a flagpole are 23° 17' and 42° 38', respectively. How tall is the pole?

214. At two points 65 feet apart on the same side of a tree and in line with it, the angles of elevation of the top of the tree are 21° 19' and 16° 20'. Find the height of the tree.

215. As a balloon passes between two points A and B, 2 miles apart, the angles of elevation of the balloon at these points are 27° 19' and 41° 45', respectively. Find the altitude of the balloon. Take A and B at the same level.

233. The top of a lighthouse is 230 feet above the sea. How far away is an object which is just “on the horizon”? [Assume the earth is a sphere of radius 3956 miles.]

234. What must be the elevation of an observer in order that he may be able to see an object on the earth thirty miles away? Assume the earth to be a smooth sphere.

In each of the figures named in the next few problems the object is to express its area (i) in terms of the radius R, that is, the radius of the circumscribed circle, (ii) in terms of the apothem r, that is, the radius of the inscribed circle, and (iii) in terms of the side a.

251. Equilateral triangle. [See problem 67 above.]
252. Square.
253. Regular pentagon.
254. Regular hexagon.
255. Regular octagon.

#### Hints

26. You only need the sin, cos, and tan of the angles A and B; you don’t need the angles themselves. So you only need the third side, which you can compute with the Pythagorean theorem, and then take ratios of two of the sides.

31. You know c and tan B. Unfortunately, tan B is the ratio of the two sides you don’t know, namely, b/a. There is more than one way to solve this problem. Here are two.

Method 1. Take the equation 1.2 = tan B = b/a, to get a relation between a and b, namely b = 1.2a. The Pythagorean theorem then gives 6.252 = a2 + 1.44 a2, from which you can determine a, and then find b.

Method 2. From tan B, you can determine the angle B (use arctan). From that you can find cos B, and then a, and you can find sin B, and then b.

34. Since you have a and b, you can use tangents to find A and the Pythagorean theorem to find c.

42. Find B by tangents and c by the Pythagorean theorem.

61. Start by drawing the figure. Although the triangle ABC is not a right triangle, it does break into two right triangles. You can use tangents to find the two parts of the side AB and add them together.

67. An equilateral triangle ABC has three 60° vertex angles. Drop a perpendicular from one vertex, say vertex C, and you get two congruent right triangles ACF and BCF, and you can find the length of that perpendicular, and that’s the altitude of the equilateral triangle. The circumscribed circle is the one passing through the three vertices, and the inscribed circle is the one inside tangent to the three sides. By dropping perpendiculars from the another of the vertices of the equilateral triangle and using trig on the resulting small triangles, you can find the radii of these two circles.

202. Since you know the height of your building and angle of depression to the base of the other building, you can determine how far away it is. Then the angle of elevation to the top of the other building will tell you how much higher it is than yours.

207. Similar hint to 202. See, trig can be useful if you’re a lonely lighthouse keeper and don’t know what to do!

214. This is a useful problem. You can use it to find heights of inaccessible things. Draw the figure. There are two unknowns: the height y of the tree, and the distance x of the nearer point to the tree. The further point is then x + 65 feet from the tree. Using tangents of the known angles, you can set up two equations which can be solved to determine y and x.

215. This is similar to 214, but in this problem, the balloon lies between the two points. Draw the figure. Decide on your variables. Set up equations, and solve them.

233. A very interesting problem. Various inverses of it have been used for centuries to compute the radius of the earth. In this problem we get to assume we know about the earth. All you need here is the Pythagorean theorem. One side of a right triangle is r, the radius of the earth, and the hypotenuse is r + h where h is the height of the lighthouse. The Pythagorean theorem the third side of the triangle.

234. Set this problem up similar to 233, but different variables are known.

251–255. You might do these all at once, saving the computations for last. Let n be the number of sides on the regular polygon. Draw lines from the center of the figure to the vertices and to the midpoints of of the sides. You get 2n little triangles. Each one of these is a right triangle with hypotenuse R, one leg r, and the other leg a/2. The angle at the center is 360°/(2n) = 180°/n. Using trigonometry, you can easily write equations relating the area of the regular polygon as required.

26. (i). b = 40. So sin A = cos B = 9/41, cos A = sin B = 40/41, tan A = 9/40, tan B = 40/9.
(ii). b = 12. So sin A = cos B = 35/37, cos A = sin B = 12/37, tan A = 35/12, tan B = 12/37.
(iii). c = 25. So sin A = cos B = 24/25, cos A = sin B = 7/25, tan A = 24/7, tan B = 7/24.

31. a = 4 feet, b = 4.8 feet, about 4'10".

34. A = 27.55°, about 28°. c = 2.6.

42. B = 26.95°, or 26°57'. c = 11.3.

61. AB = 12/tan A + 12/tan B = 12(√3 + 1) inches, about 33".

67. (a√3)/2, (a√3)/3, and (a√3)/6, respectively.

202. Distance = 50/tan 26°34' = 100 feet. Height = 50 + 100 tan 19°41' = 85.8' = 85'9".

207. Distance = 175/tan 42°38' = 190 feet. Height = 175 - 190 tan 23°17' = 93.23' = 9'3".

214. The two equations are

0.293052 = tan 16°20' = h/(65 + x), and
0.390219 = tan 21°19' = h/x.
where x is the distance from the nearest point to the base of the tree. You can solve these simultaneously for x and h.
Distance x = 196'. Height h = 76.5'.

215. If h is the height of the balloon, and x is the distance along the ground from A to the point directly under the balloon, then the two equarions are

tan 27°19' = h/x, and
tan 41°45' = h/(2 – x)
You can solve this pair of equations for x and h.
Height = .654 miles = 3455 feet.

233. A trifle more than 18.5 miles.

234. 600 feet.

251–255. The area of a regular n-gon is A = nra/2. To find A in terms of R, r, or a, use the relationships

cos 180°/n = r/R, and
tan 180°/n = a/(2r).
Then
(i) in terms of R, the area A = nR2 cos 180°/n sin 180°/n,
(ii) in terms of r, the area A = nr2 tan 180°/n, and
(iii) in terms of a, the area A = na2/(4tan 180°/n).

Problemshape(i) R(ii) r(iii) a
251triangle (3R2 √3)/4 3r2 √3 (a2 √3)/4
252square 2R2 4r2 a2
253pentagon (5R2 sin 108°)/2 5r2 tan 36° (5a2 tan 54°)/4
254hexagon (3R2 √3)/2 2r2 √3 (3a2 √3)/2
255octagon 2R2 √2 8r2 tan 22°30' 2a2 tan 67°30'

#### An aside on Pythagorean triples

This doesn’t have much to do with trigonometry, but it’s interesting. You’ve probably noticed how Crowley often chose two sides of a right triangle to be whole numbers, and the third turns out to be a whole number, too. Like in problem 26 where all three right triangles had whole numbers for sides, namely 9:40:41, 12:35:37, and 7:24:25. Also, at the beginning of this page there was a 5:12:13 triangle (actually 10:24:26, but that’s similar to a 5:12:13 triangle). And no doubt you already know about the 3:4:5 right triangle.

So, are there other special right triangles whose sides are all whole numbers? Yes, and they’ve been studied for a long time. Three numbers a, b, and c such that a2 + b2 = c2 are said to form a Pythagorean triple, in honor of Pythagoras. He lived about 550 B.C.E. and probably know quite a few of them. But the Old Babylonians of about 1800 B.C.E. knew them all, and many were known in other ancient civilizations such as China and India.

Before reading the paragraph, see if you can find some more Pythagorean triples. Don’t count any that have a common factor as new, such as 6:8:10, since they’ll be similar to smaller ones.

In Euclid’s Elements there is a description of all the possible Pythagorean triples. Here’s a modern paraphrase of Euclid. Take any two odd numbers m and n, with m <  n, and relatively prime (that is, no common factors). Let a = mn, let b = (n2 – m2)/2, and let c = (n2 + m2)/2. Then a:b:c is a Pythagorean triple. For instance, if you take m = 1, and n = 3, then you get the smallest Pythagorean triple 3:4:5.