Hungerford T.W., Shaw D.J. Contemporary Precalculus: A Graphing Approach
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40. A 5-inch-high plastic beverage glass has a 2.5-inch-
diameter base. Its sides slope outward at a 4° angle as
shown. What is the diameter of the top of the glass?
41. In aerial navigation, directions are given in degrees clock-
wise from north. Thus, east is 90°, south is 180°, and so on,
as shown in the figure. A plane travels from an airport for
200 miles in the direction 300°. How far west of the airport
is the plane then?
42. A plane travels at a constant 300 mph in the direction 65°
(see Exercise 41).
(a) How far east of its starting point is the plane after half
an hour?
(b) How far north of its starting point is the plane after
2 hours and 24 minutes?
43. A closed 60-foot-long drawbridge is 24 feet above water
level. When open, the bridge makes an angle of 33° with the
horizontal.
(a) How high is the tip P of the open bridge above the
water?
(b) When the bridge is open, what is the distance from P
to Q?
Q
P
33°
0° North
180° South
90° East270° West
300°
Distance west
of airport
4° 4°
596 CHAPTER 8 Triangle Trigonometry
THINKERS
44. A gutter is to be made from a strip of metal 24 inches wide
by bending up the sides to form a trapezoid.
(a) Express the area of the cross section of the gutter as a
function of the angle t. [Hint: The area of a trapezoid
with bases b and b and height h is h(b b)/2.]
(b) For what value of t will this area be as large as possible?
45. The cross section of a tunnel is a semicircle with radius
10 meters. The interior walls of the tunnel form a rectangle.
(a) Express the area of the rectangular cross section of the
tunnel opening as a function of angle t.
(b) For what value of t is the cross-sectional area of the
tunnel opening as large as possible? What are the di-
mensions of the tunnel opening in this case?
46. A spy plane on a practice run over the Midwest takes a pic-
ture that shows Cleveland, Ohio, on the eastern horizon
and St. Louis, Missouri, 520 miles away, on the western
horizon (the figure is not to scale). Assuming that the ra-
dius of the earth is 3950 miles, how high was the plane
when the picture was taken? [Hint: The sight lines from
the plane to the horizons are tangent to the earth, and a
tangent line to a circle is perpendicular to the radius at
that point. The arc of the earth between St. Louis and
Cleveland is 520 miles long. Use this fact and the arc
y
t
10 10
x
8
8
8
h
t
length formula to find angle u (your answers will be in
radians). Note that a u/2 (why?).]
α
θ
St. Louis Cleveland
SECTION 8.3 The Law of Cosines 597
47. A 50-foot-high flagpole stands on top of a building. From a
point on the ground, the angle of elevation of the top of the
pole is 43°, and the angle of elevation of the bottom of
the pole is 40°. How high is the building?
48. Two points on level ground are 500 meters apart. The angles
of elevation from these points to the top of a nearby hill are
52° and 67°, respectively. The two points and the ground
level point directly below the top of the hill lie on a straight
line. How high is the hill?
8.3 The Law of Cosines
■ Use the Law of Cosines to solve oblique triangles.
■ Use the Law of Cosines to solve applied problems.
We now consider the solution of oblique triangles (ones that don’t contain a right
angle). We shall use standard notation for triangles: Each vertex is labeled with
a capital letter, and the length of the side opposite that vertex is denoted by the
same letter in lower case, as shown in Figure 8–39. The letter A will also be used
to label the angle at vertex A and similarly for B and C. So we shall make state-
ments such as A 37° or cos B .326.
The first fact needed to solve oblique triangles is the Law of Cosines, whose
proof is given at the end of this section.
Section Objectives
Law of
Cosines
In any triangle ABC, with sides of lengths a, b, c, as in Figure 8–39,
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
You need only memorize one of these equations since each of them provides
essentially the same information: a description of one side of a triangle in terms of
the angle opposite it and the other two sides.
NOTE
When C is a right angle, then c is the hypotenuse and
cos C cos 90° 0.
In this case, the third equation in the Law of Cosines becomes
the Pythagorean Theorem:
c
2
a
2
b
2
.
So the Pythagorean Theorem is a special case of the Law of Cosines.
C
A
B
b
c
a
Figure 8–39
C
A
B
b
c
a
EXAMPLE 1
If the triangle in Figure 8–39 has a 7, c 15 and B 60°, find b.
SOLUTION Using the second equation in the Law of Cosines, we have
b
2
a
2
c
2
2ac cos B
7
2
15
2
2
7
15 cos 60°
49 225 210 cos 60°
274 210
1
2
b
2
169
Hence, b
169
13. ■
Sometimes it is more convenient to use the Law of Cosines in a slightly different
form. To do this we solve the first equation in the Law of Cosines for cos A.
a
2
b
2
c
2
2bc cos A
Add 2bc cos A to both sides: 2bc cos A a
2
b
2
c
2
Subtract a
2
from both sides: 2bc cos A b
2
c
2
a
2
Divide both sides by 2bc: cos A
b
2
2
c
b
2
c
a
2
So we have the following result.
598 CHAPTER 8 Triangle Trigonometry
Law of Cosines:
Alternate Form
In any triangle ABC, with sides of lengths a, b, c, as in Figure 8–39,
cos A
b
2
2
c
b
2
c
a
2
.
The other two equations can be similarly rewritten. In this form, the Law of Cosines
provides a description of each angle of a triangle in terms of the three sides. Conse-
quently, the Law of Cosines can be used to solve triangles in these cases:
1. Two sides and the angle between them are known (SAS).
2. Three sides are known (SSS).
EXAMPLE 2
SAS Solve triangle ABC in Figure 8–40.
SOLUTION We have a 16, b 10, and C 110°. The right side of the third
equation in the Law of Cosines involves only these known quantities. Hence,
c
2
a
2
b
2
2ab cos C
c
2
16
2
10
2
2
16
10 cos 110°
c
2
256 100 320(.342) 465.4.*
10 16
A
B
c
C
110°
Figure 8–40
*Throughout this chapter, all decimals are printed in rounded-off form for reading convenience, but no
rounding is done in the actual computation until the final answer is obtained.
Therefore, c
465.4
21.6. Now use the alternate form of the Law of
Cosines.
cos A
b
2
2
c
b
2
c
a
2
.7172.
A calculator (in degree mode) shows that cos
1
(.7172) 44.2°. So A 44.2° is
an angle with cosine .7172. Hence,
B 180° A C
180° 44.2° 110° 25.8°. ■
EXAMPLE 3
SSS Find the angles of triangle ABC in Figure 8–41.
SOLUTION In this case, a 20, b 15, and c 8.3. By the alternate form of
the Law of Cosines,
cos A
b
2
2
c
b
2
c
a
2
15
2
2
1
8
5
.3
2
8.3
20
2
1
2
0
4
6
9
.11
.4261.
The COS
1
key shows that A 115.2°. Similarly, the alternate form of the Law
of Cosines yields
cos B
a
2
2
c
a
2
c
b
2
cos B
20
2
2
2
8
0
.3
2
8.3
15
2
24
3
3
3
.
2
89
.7346
B 42.7°.
Therefore, C 180° 115.2° 42.7° 22.1°. ■
EXAMPLE 4
Two trains leave a station on different tracks. The tracks make an angle of 125°
with the station as vertex. The first train travels at an average speed of 100 kilo-
meters per hour, and the second travels at an average speed of 65 kilometers per
hour. How far apart are the trains after 2 hours?
10
2
(21.6)
2
16
2
2
10
21.6
SECTION 8.3 The Law of Cosines 599
15
8.3
C
B20
A
Figure 8–41
SOLUTION The first train A traveling at 100 kilometers per hour for 2 hours
goes a distance of 100 2 200 kilometers. The second train B travels a distance
of 65 2 130 kilometers. So we have the situation shown in Figure 8–42.
By the Law of Cosines,
c
2
a
2
b
2
2ab cos C
130
2
200
2
2
130
200 cos 125°
56,900 52,000 cos 125° 86,725.97
c 86,725
.97
294.5 kilometers.
The trains are 294.5 kilometers apart after 2 hours. ■
EXAMPLE 5
A small powerboat leaves Chicago and sails 35 miles due east on Lake Michigan.
It then changes course 59° northward, heading for Grand Haven, Michigan, as
shown in Figure 8–43. After traveling 60 miles on this course, how far is the boat
from Chicago?
600 CHAPTER 8 Triangle Trigonometry
Chicago
Grand
Haven
35
60
59
A
B
C
Figure 8–43
SOLUTION Figure 8–43 shows that
B 59° 180°
B 180° 59° 121°.
We must find the side of the triangle opposite angle B. By the Law of Cosines,
b
2
a
2
c
2
2ac cos B
b
2
35
2
60
2
2
35
60 cos 121°
b
2
6988.1599
b 6988.1
599
83.5952.
So the boat is about 83.6 miles from Chicago. ■
125°
C
B
A200
130
Station
c
Figure 8–42
EXAMPLE 6
A sculpture is being placed in front of a new office building. The sculpture con-
sists of two steel beams of lengths 10 and 12 feet, respectively, and a 15.2-foot
cable, as shown in Figure 8–44. If the 10-foot beam makes an angle of 50° with
the ground, what angle does the 12-foot beam make with the ground?
SECTION 8.3 The Law of Cosines 601
A
50°
10
12
15.2
Figure 8–44
SOLUTION We must find the measure of angle u. As you can see in Fig-
ure 8–44,
50° A u 180°.
We first find the measure of angle A and then solve this equation for u. The triangle
formed by the beams and cable has sides of lengths 10, 12, and 15.2, with angle A
opposite the 15.2-foot side. By the alternate form of the Law of Cosines,
cos A
10
2
2
1
1
2
0
2
12
15.2
2
.054.
Hence, the measure of angle A is about 86.9° (Figure 8–45). Therefore,
50° A u 180°
50° 86.9° u 180°
u 180° 50° 86.9° 43.1°. ■
EXAMPLE 7
A 100-foot-tall antenna tower is to be placed on a hillside that makes an angle of
12° with the horizontal. It is to be anchored by two cables from the top of the
tower to points 85 feet uphill and 95 feet downhill from the base. How much cable
is needed?
SOLUTION The situation is shown in Figure 8–46 on the next page.
Figure 8–45
In triangle BEC, angle E is a right angle and by hypothesis, angle C measures 12°.
Since the sum of the angles of a triangle is 180°, we must have
⭿CBE 180° (90° 12°) 78°.
As shown in the figure, the sum of angles CBE and CBA is a straight angle (180°).
Hence,
⭿CBA 180° 78° 102°.
Apply the Law of Cosines to triangle ABC.
b
2
a
2
c
2
2ac cos B
b
2
95
2
100
2
2
95
100 cos 102°
9025 10,000 19,000 cos 102°
22,975.32.
Therefore, the length of the downhill cable is b 22,975
.32
151.58 feet.
To find the length of the uphill cable, note that the sum of angles CBA and
DBA is a straight angle, so
⭿DBA 180° ⭿CBA 180° 102° 78°.
Applying the Law of Cosines to triangle DBA, we have
k
2
85
2
100
2
2
85
100 cos 78°
7225 10,000 17,000 cos 78° 13,690.50.
Hence, the length of the uphill cable is k 13,690
.50
117.01 feet. ■
PROOF OF THE LAW OF COSINES
Given triangle ABC, position it on a coordinate plane so that angle A is in standard
position with initial side c and terminal side b. Depending on the size of angle A,
there are two possibilities, as shown in Figure 8–47.
602 CHAPTER 8 Triangle Trigonometry
12°
D
A
b
k
C
B
E
100
95
85
Figure 8–46
The coordinates of B are (c, 0). Let (x, y) be the coordinates of C. Now C is a point on
the terminal side of angle A, and the distance from C to the origin A is obviously b.
Therefore, according to the point-in-the-plane description of sine and cosine, we have
b
x
cos A or, equivalently, x b cos A,
b
y
sin A or, equivalently, y b sin A.
Using the distance formula on the coordinates of B and C, we have
a distance from C to B
(x c
)
2
(
y 0)
2
(b cos
A c)
2
(b
sin A
0)
2
.
Squaring both sides of this last equation and simplifying, using the Pythagorean
identity, yields
a
2
(b cos A c)
2
(b sin A)
2
a
2
b
2
cos
2
A 2bc cos A c
2
b
2
sin
2
A
a
2
b
2
(sin
2
A cos
2
A) c
2
2bc cos A
a
2
b
2
c
2
2bc cos A.
This proves the first equation in the Law of Cosines. Similar arguments beginning
with angle B or C in standard position prove the other two equations.
SECTION 8.3 The Law of Cosines 603
BA
c
C(x, y)
a
b
BA
c
C(x, y)
a
b
Figure 8–47
EXERCISES 8.3
Directions: Standard notation for triangle ABC is used
throughout. Use a calculator and round off your answers to
one decimal place at the end of the computation.
In Exercises 1–16, solve the triangle ABC under the given
conditions.
1. A 40°, b 10, c 7
2. B 40°, a 12, c 20
3. C 118°, a 6, b 12
4. C 52.5°, a 6.5, b 9
5. A 140°, b 12, c 14
6. B 25.4°, a 6.8, c 10.5
7. C 78.6°, a 12.1, b 20.3
8. A 118.2°, b 16.5, c 10.7
9. a 7, b 3, c 5
10. a 8, b 5, c 10
11. a 16, b 30, c 32
12. a 5.3, b 7.2, c 10
13. a 7.2, b 6.5, c 11
14. a 6.8, b 12.4, c 15.1
15. a 12, b 16.5, c 20.6
16. a 5.7, b 20.4, c 16.8
17. Find the angles of the triangle whose vertices are (0, 0),
(5, 2), (1, 4).
18. Find the angles of the triangle whose vertices are (3, 4),
(6, 1), (2, 1).
19. In Example 4, suppose that the angle between the two tracks
is 112° and that the average speeds are 90 kilometers per
hour for the first train and 55 kilometers per hour for the
second train. How far apart are the trains after two hours
and 45 minutes?
20. Suppose that the boat in Example 5 goes 25 miles due east
and then changes course 56° northward. After traveling
50 miles on this course, how far is the boat from Chicago?
21. The sculptor builds a smaller version of the sculpture in Ex-
ample 6, in which the beams are six feet and nine feet long,
respectively, and the cable is 10.4 feet long. If the six-foot
beam makes an angle of 40° with the ground, what angle
does the nine-foot beam make with the ground?
22. Suppose that the tower in Example 7 is 175 feet high and
that the cable on the downhill side is 120 feet from the base
of the tower. How long is that cable?
23. The pitcher’s mound on a standard baseball diamond (which
is actually a square) is 60.5 feet from home plate (see the fig-
ure). How far is the pitcher’s mound from first base?
24. At Wrigley Field in Chicago, the straight-line distance from
home plate over second base to the center field wall is 400
feet. How far is it from first base to the same point at the
center field wall? [Hint: Adapt and extend the figure from
Exercise 23.]
25. A stake is located 10.8 feet from the end of a closed gate that
is 8 feet long. The gate swings open, and its end hits the
stake. Through what angle did the gate swing?
8
10.8
Home plate
90 ft
90 ft
60.5 ft
90 ft90 ft
1st base
3rd base
2nd base
Pitcher's
mound
604 CHAPTER 8 Triangle Trigonometry
26. The distance from Chicago to St. Louis is 440 kilometers,
that from St. Louis to Atlanta 795 kilometers, and that from
Atlanta to Chicago 950 kilometers. What are the angles in
the triangle with these three cities as vertices?
27. A satellite is placed in an orbit such that the satellite remains
stationary 24,000 miles over a fixed point on the surface of
the earth. The angle CES, where C is Cape Canaveral, E is
the center of the earth, and S is the satellite, measures 60°.
Assuming that the radius of the earth is 3960 miles, how far
is the satellite from Cape Canaveral?
28. One plane flies west from Cleveland at 350 mph. A second
plane leaves Cleveland at the same time and flies southeast
at 200 mph. How far apart are the planes after 1 hour and
36 minutes?
29. A weight is hung by two cables from a beam. What angles
do the cables make with the beam?
30. Two ships leave port, one traveling in a straight course at
22 mph and the other traveling a straight course at 31 mph.
Their courses diverge by 38°. How far apart are they after
3 hours?
31. A boat runs in a straight line for 3 kilometers, then makes a
45° turn and goes for another 6 kilometers (see the figure).
How far is the boat from its starting point?
32. A plane flies in a straight line at 400 mph for 1 hour and
12 minutes. It makes a 15° turn and flies at 375 mph for
2 hours and 27 minutes. How far is it from its starting point?
45°
Start
3
6
50 ft
?
?
100 ft
65 ft
S
C
E
33. A surveyor wants to measure the width CD of a sinkhole.
So he places a stake B and determines the measurements
shown in the figure. How wide is the sinkhole?
34. A straight tunnel is to be dug through a hill. Two people
stand on opposite sides of the hill where the tunnel en-
trances are to be located. Both can see a stake located
530 meters from the first person and 755 meters from the
second. The angle determined by the two people and the
stake (vertex) is 77°. How long must the tunnel be?
35. A 400-foot-high tower stands on level ground, anchored by
two cables on the west side. The end of the cable closest to
the tower makes an angle of 70° with the horizontal. The
two cable ends are 100 feet apart, as shown in the figure.
How long are the cables?
36. One diagonal of a parallelogram is 6 centimeters long, and
the other is 13 centimeters long. They form an angle of 42°
with each other. How long are the sides of the parallelo-
gram? [Hint: The diagonals of a parallelogram bisect each
other.]
37. A ship is traveling at 18 mph from Corsica to Barcelona, a
distance of 350 miles. To avoid bad weather, the ship leaves
Corsica on a route 22° south of the direct route (see the
figure). After 7 hours, the bad weather has been bypassed.
Through what angle should the ship now turn to head
directly to Barcelona?
22°
Angle of
turn
Barcelona
Corsica
70°
100 ft
400 ft
103°
C
D
B
120 ft
74 ft
SECTION 8.3 The Law of Cosines 605
38. A plane leaves South Bend for Buffalo, 400 miles away,
intending to fly a straight course in the direction 70° (aerial
navigation is explained in Exercise 41 of Section 8.2). After
flying 180 miles, the pilot realizes that an error has been
made and that he has actually been flying in the direc-
tion 55°.
(a) At that time, how far is the plane from Buffalo?
(b) In what direction should the plane now go to reach
Buffalo?
39. Assume that the earth is a sphere of radius 3960 miles.
A satellite travels in a circular orbit around the earth,
900 miles above the equator, making one full orbit every
6 hours. If it passes directly over a tracking station at 2
P.M.,
what is the distance from the satellite to the tracking station
at 2:05
P.M.?
40. A surveyor has determined the distance and angles in the
figure. He wants you to find the straight-line distance from
A to B. Do so.
41. A parallelogram has diagonals of lengths 12 and 15 inches
that intersect at an angle of 63.7°. How long are the sides of
the parallelogram? [See the hint for Exercise 36.]
42. Two planes at the same altitude approach an airport. One
plane is 16 miles from the control tower and the other is
22 miles from the tower. The angle determined by the
planes and the tower, with the tower as vertex, is 11°. How
far apart are the planes?
43. Assuming that the circles in the figure are mutually tangent,
find the lengths of the sides and the measures of the angles
in triangle ABC.
C
A
B
13
8.23
11.27
3000 ft
3300 ft
1400 ft
2000 ft
1200 ft
52
75
A
C
B