Let’s agree again to the standard convention for labelling the parts of a right triangle. Let the right angle be labelled

- Pythagorean theorem:
*a*^{2}+*b*^{2}=*c*^{2}. - 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.*

Now suppose we know the hypotenuse and one side, but have to find the other. For example, if *b* = 119 and *c* = 169, then *a*^{2} = *c*^{2} – *b*^{2} = 169^{2} – 119^{2} = 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.

That’s all there is to it.

- sin
*A*=*a/c*(opp/hyp) - cos
*A*=*b/c*(adj/hyp) - tan
*A*=*a/b*(opp/adj) - cot
*A*=*b/a*(adj/opp) - sec
*A*=*c/b*(hyp/adj) - csc
*A*=*c/a*(hyp/opp)

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.*

**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.

**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* = 1.2*a.* The Pythagorean theorem then gives 6.25^{2} = *a*^{2} + 1.44 *a*^{2}, 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 2*n* 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°/(2*n*) = 180°/*n.* Using trigonometry, you can easily write equations relating the area of the regular polygon as required.

(ii).

(iii).

**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' =

0.390219 = tan 21°19' =

Distance

**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' =

tan 41°45' =

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°/

tan 180°/

- (i) in terms of

(ii) in terms of

(iii) in terms of

Problem | shape | (i) R | (ii) r | (iii) a |
---|---|---|---|---|

251 | triangle | (3R^{2} √3)/4 |
3r^{2} √3 |
(a^{2} √3)/4 |

252 | square | 2R^{2} |
4r^{2} |
a^{2} |

253 | pentagon | (5R^{2} sin 108°)/2 |
5r^{2} tan 36° |
(5a^{2} tan 54°)/4 |

254 | hexagon | (3R^{2} √3)/2 |
2r^{2} √3 |
(3a^{2} √3)/2 |

255 | octagon | 2R^{2} √2 |
8r^{2} tan 22°30' |
2a^{2} tan 67°30' |

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 *a*^{2} + *b*^{2} = *c*^{2} 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* = (*n*^{2} – *m*^{2})/2, and let *c* = (*n*^{2} + *m*^{2})/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.