Trigonometry means “measurement of triangles.” Consider its parts: tri means “three,” gon means “angle,” and metry means “measure.” Thus, in trigonometry we study the measurement of triangles.
The following ratios relate the sides and acute angles of a right triangle:
1. Tangent ratio: The tangent (abbreviated “tan”) of an acute angle equals the length of the leg opposite the angle divided by the length of the leg adjacent to the angle.
2. Sine ratio: The sine (abbreviated “sin”) of an acute angle equals the length of the leg opposite the angle divided by the length of the hypotenuse.
3. Cosine ratio: The cosine (abbreviated “cos”) of an acute angle equals the length of the leg adjacent to the angle divided by the length of the hypotenuse.
Thus in right triangle ABC of Fig. 8-1,
Fig. 8-1
If A and B are the acute angles of a right triangle, then
A scientific calculator can compute the sine, cosine, and tangent of an angle with the “SIN,” “COS,” and “TAN” buttons, respectively. Make sure the calculator is set to degrees (DEG). For those without a calculator, a table of sines, cosines, and tangents is in the back of this book.
8.1 Using the table of sines, cosines, and tangents
The following values were taken from a table of sines, cosines, and tangents. State, in equation form, what the values on the first three lines mean. Then use the table at the back of this book to complete the last line.
Solutions
(a) sin 1° = 0.0175; cos 1° = 0.9998; tan 1° = 0.0175
(b) sin 30° = 0.5000; cos 30° = 0.8660; tan 30° = 0.5774
(c) sin 60° = 0.8660; cos 60° = 0.5000; tan 60° = 1.7321
(d) In the table of trigonometric functions, the cosine value 0.3420 is on the 70° line; hence, the angle measures 70°. Then, from the table, sin 70° = 0.9397 and tan 70° = 2.7475.
8.2 Finding angle measures to the nearest degree
Find the measure of x to the nearest degree if (a) sin x = 0.9235; (b) cos x = 
or 0.8400; (c) tan x = 
/ 10 or 0.2236. Use the table of trigonometric functions.
Solutions
With a calculator, the above answers can be found with the inverse sine (sin–1), inverse cosine (cos–1), and inverse tangent (tan–1). These usually require first pressing the “2nd” or “INV” button and then “SIN,” “COS,” or “TAN.”
8.3 Finding trigonometric ratios
For each right triangle in Fig. 8-2, find the trigonometric ratios of each acute angle.
Fig. 8-2
Solutions
8.4 Finding measures of angles by trigonometric ratios
Find the measure of angle A, to the nearest degree, in each part of Fig. 8-3.
Fig. 8-3
Solutions
(a) 
Since sin 37° = 0.6018 is the nearest-degree sine value, m∠A = 37°.
(b) 
Since tan 23° = 0.4245 is the nearest-degree tangent value, m∠A = 23°.
Since cos 28° = 0.8829 is the nearest-degree cosine value, m∠A = 28°.
8.5 Trigonometric ratios of 30° and 60°
Show that
(a) tan 30° = 0.577
(b) sin 30° = 0.500
(c) cos 30° = 0.866
(d) tan 60° = 1.732
(e) sin 60° = 0.866
(f) cos 60° = 0.500
Fig. 8-4
Solutions
The trigonometric ratios for 30° and 60° may be obtained by using a 30°-60°-90° triangle (Fig. 8-4); in such a triangle, the ratio of the sides is a:b:c = 1: 
:2. Thus:
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
8.6 Finding lengths of sides by trigonometric ratios
In each triangle of Fig. 8-5, solve for x and y to the nearest integer.
Fig. 8-5
Solutions
(a) Since tan 40° = 
, x = 150 tan 40° = 150(0.8391) = 126.
Since cos 40° 
(b) Since tan 50° = , x = 150 tan 50° = 150(1.1918) = 179.
Since sin 40° 
(c) Since sin 40° = , x = 150 sin 40° = 150 (0.6428) = 96.
Since cos 40° = 
, y = 150 cos 40° = 150(0.776) = 115.
8.7 Solving trigonometry problems
(a) An aviator flew 70 mi east from A to C. From C, he flew 100 mi north to B. Find the measure of the angle of the turn (to the nearest degree) that must be made at B to return to A.
(b) A road is to be constructed so that it will rise 105 ft for each 1000 ft of horizontal distance. Find the measure of the angle of rise to the nearest degree, and the length of road to the nearest foot for each 1000 ft of horizontal distance.
(a) The required angle is ∠EBA in Fig. 8-6(a). In rt. ΔABC, tan 
= 0.7000; hence, m∠B = 35° and m∠ EBA = 180°–35° = 145°.
(b) We need to find m∠A and x in Fig. 8-6(b). Since tan 
= 0.1050, m∠A = 6°. Then 
so 
Fig. 8-6
Here are some definitions that are involved in observed-angle problems:
The line of sight is the line from the eye of the observer to the object sighted.
A horizontal line is a line that is parallel to the surface of water.
An angle of elevation (or depression) is an angle formed by a horizontal line and a line of sight above (or below) the horizontal line and in the same vertical plane.
Thus in Fig. 8-7, the observer is sighting an airplane above the horizontal, and the angle formed by the horizontal and the line of sight is an angle of elevation. In sighting the car, the angle her line of sight makes with the horizontal is an angle of depression.
Fig. 8-7
8.8 Using an angle of elevation
(a) Sighting to the top of a building, Henry found the angle of elevation to measure 21°. The ground is level. The transit is 5 ft above the ground and 200 ft from the building. Find the height of the building to the nearest foot.
(b) If the angle of elevation of the sun at a certain time measures 42°, find to the nearest foot the height of a tree whose shadow is 25 ft long.
(a) If x is the height of the part of the building above the transit [Fig. 8-8(a)], then 
and x = 200 tan 21° = 200(0.3839) = 77 ft.
Thus, the height of the building is h = x + 5 = 77 + 5 = 82 ft.
(b) If h is the height of the tree [Fig. 8-8(b)], then we have 
and h = 25 tan 42° = 25(0.9004) = 23 ft.
Fig. 8-8
8.9 Using both an angle of elevation and an angle of depression
Standing at the top of a lighthouse 200 ft high, a lighthouse keeper sighted both an airplane and a ship directly beneath the plane. The angle of elevation of the plane measured 25°; the angle of depression of the ship measured 32°. Find (a) the distance d of the boat from the foot of the lighthouse, to the nearest 10 ft; (b) the height of the plane above the water, to the nearest 10 ft.
Solutions
(a) See Fig. 8-9. In ΔIII, 
and d = 200 tan 58° = 200(1.6003) = 320 ft.
(b) In ΔI, 
and x = 320(0.4663) = 150 ft. Since the height of the tower is 200 ft, the height of the airplane is 200 + 150 = 350 ft.
Fig. 8-9
8.10 Using two angles of depression
An observer on the top of a hill 250 ft above the level of a lake sighted two boats directly in line. Find, to the nearest foot, the distance between the boats if the angles of depression noted by the observer measured 11° and 16°.
In ΔAB'C of Fig. 8-10, m∠B'AC = 90°–11° = 79. Then CB' = 250 tan 79°.
In ΔABC, m∠BAC = 90°–16° = 74°. Then CB = 250 tan 74°.
Hence, BB' = CB'–CB = 250(tan 79°–tan 74°) = 250(5.1446–3.4874) = 414 ft.
Fig. 8-10
8.1. Using the table of trigonometric functions at the end of the book, find
(8.1)
(a) sin 25°, sin 48°, sin 59°, and sin 89°
(b) cos 15°, cos 52°, cos 74°, and cos 88°
(c) tan 4°, tan 34°, tan 55°, and tan 87°
(d) which trigonometric ratios increase as the measure of the angle increases from 0° to 90°
(e) which trigonometric ratio decreases as the measure of the angle increases from 0° to 90°
(f) which trigonometric ratio has values greater than 1
8.2. Using the table at the end of the book, find the angle for which
(8.1)
(a) sin x = 0.3420
(b) sin A = 0.4848
(c) sin B = 0.9455
(d) cos A' = 0.9336
(e) cos y = 0.7071
(f) cos Q = 0.3584
(g) tan W = 0.3443
(h) tan B' = 2.3559
8.3. Using the table at the end of the book, find the measure of x to the nearest degree if
(8.2)
(a) sin x = 0.4400
(b) sin x = 0.7280
(c) sin x = 0.9365
(d) cos x = 0.9900
(e) cos x = 0.7650
(f) cos x = 0.2675
(g) tan x = 0.1245
(h) tan x = 0.5200
(i) tan x = 5.5745
(j) 
(k) 
(l) 
(m) 
(n) 
(o) 
(p) 
8.4. In each right triangle of Fig. 8-11, find sin A, cos A, and tan A.
(8.3)
Fig. 8-11
8.5. Find m∠A to the nearest degree in each part of Fig. 8-12.
(8.4)
Fig. 8-12
8.6. Find m∠B to the nearest degree if (a) b = 67 and c = 100; (b) a = 14 and c = 50; (c) a = 22 and b = 55; (d) a = 3 and b = in Fig. 8-13.
(8.4)
Fig. 8-13
Fig. 8-14
8.7. Making use of a square with a side of 1 (Fig. 8-14), show that (a) the diagonal c = 
; (b) tan 45° = 1; (c) sin 45° = cos 45° = 0.707.
(8.5)
8.8. To the nearest degree, find the measure of each acute angle of any right triangle whose sides are in the ratio of (a) 5:12:13; (b) 8:15:17; (c) 7:24:25; (d) 11:60:61.
(8.4)
8.9. In each triangle of Fig. 8-15, solve for x and y to the nearest integer.
(8.5)
Fig. 8-15
8.10. A ladder leans against the side of a building and makes an angle measuring 70° with the ground. The foot of the ladder is 30 ft from the building. Find, to the nearest foot, (a) how high up on the building the ladder reaches; (b) the length of the ladder.
(8.6)
8.11. To find the distance across a swamp, a surveyor took measurements as shown in Fig. 8-16. 
is at right angles to 
. If m∠A = 24° and AC = 350 ft, find the distance BC across the swamp.
(8.6)
Fig. 8-16
Fig. 8-17
8.12. A plane rises from take-off and flies at a fixed angle measuring 9° with the horizontal ground (Fig. 8-17). When it has gained 400 ft in altitude, find, to the nearest 10 ft, (a) the horizontal distance flown; (b) the distance the plane has actually flown.
(8.6)
8.13. The base angle of an isosceles triangle measures 28°, and each leg has length 45 in (Fig. 8-18). Find, to the nearest inch, (a) the length of the altitude drawn to the base; (b) the length of the base.
(8.6)
Fig. 8-18
8.14. In a triangle, an angle measuring 50° is included between sides of length 12 and 18. Find the length of the altitude to the side of length 12, to the nearest integer.
(8.6)
8.15. Find the lengths of the sides of a rectangle to the nearest inch if a diagonal of length 24 in makes an angle measuring 42° with a side.
(8.6)
8.16. A rhombus has an angle measuring 76° and a long diagonal of length 40 ft. Find the length of the short diagonal to the nearest foot.
(8.6)
8.17. Find the length of the altitude to the base of an isosceles triangle to the nearest yard if its base has length 40 yd and its vertex angle measures 106°.
(8.6)
8.18. A road is inclined uniformly at an angle measuring 6° with the horizontal (Fig. 8-19). After a car is driven 10,000 ft along this road, find, to the nearest 10 ft, the (a) increase in the altitude of the car and driver; (b) horizontal distance that has been driven.
(8.7)
Fig. 8-19
8.19. An airplane travels 15,000 ft through the air at a uniform angle of climb, thereby gaining 1900 ft in altitude. Find its angle of climb.
(8.7)
8.20. Sighting to the top of a monument. William found the angle of elevation to measure 16° (Fig. 8-20). The ground is level, and the transit is 5 ft above the ground. If the monument is 86 ft high, find, to the nearest foot, the distance from William to the foot of the monument.
(8.8)
Fig. 8-20
8.21. Find to the nearest degree the measure of the angle of elevation of the sun when a tree 60 ft high casts a shadow of (a) 10 ft; (b) 60 ft.
8.22. At a certain time of day, the angle of elevation of the sun measures 34°. Find, to the nearest foot, the length of the shadow cast by (a) a 15-ft vertical pole; (b) a building 70 ft high.
(8.8)
8.23. A light at C is projected vertically to a cloud at B. An observer at A, 1000 ft horizontally from C, notes the angle of elevation of B. Find the height of the cloud, to the nearest foot, if m∠A = 37°.
(8.8)
8.24. A lighthouse built at sea level is 180 ft high (Fig. 8-21). From its top, the angle of depression of a buoy measures 24°. Find, to the nearest foot, the distance from the buoy to the foot of the lighthouse.
(8.8)
Fig. 8-21
8.25. An observer, on top of a hill 300 ft above the level of a lake, sighted two ships directly in line. Find, to the nearest foot, the distance between the boats if the angles of depression noted by the observer measured (a) 20° and 15°; (b) 35° and 24°; (c) 9° and 6°.
(8.10)
8.26. In Fig. 8-22, m∠A = 43°, m∠BDC = 54°, m∠ C = 90°, and DC = 170 ft. (a) Find the length of . (b) Using the result of (a), find the length of 
.
(8.10)
Fig. 8-22
Fig. 8-23
8.27. In Fig. 8-23, m∠B = 90°, m∠ACB = 58°, m∠ D = 23°, and BC = 60 ft. (a) Find the length of . (b) Using the result of part (a), find the length of 
.
8.28. Tangents 
and 
are drawn to a circle from external point P. m∠APB = 40°, and PA = 25. Find to the nearest tenth the radius of the circle.
