592 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY

SECTION 7.2 Law of Cosines and Area

Law of Cosines
PREPARE FOR THIS SECTION
Area of a Triangle
Prepare for this section by completing the following exercises. The answers can be found
Heron’s Formula on page A46.

PS1. Evaluate 2a2 + b2 - 2ab cos C for a = 10.0, b = 15.0, and C = 110.0°.
Round your result to the nearest tenth. [5.3]

PS2. Find the area of a triangle with a base of 6 inches and a height of 8.5 inches. [1.2]

PS3. Solve c2 = a2 + b2 - 2ab cos C for C. [6.5]

PS4. The semiperimeter of a triangle is defined as one-half the perimeter of the

triangle. Find the semiperimeter of a triangle with sides of 6 meters, 9 meters,
and 10 meters. [P.1]

PS5. Evaluate 1s1s - a21s - b21s - c2 for a = 3, b = 4, c = 5, and

a + b + c
s = . [P.2] c
b
2
PS6. State a relationship between the lengths a, b, and c in the triangle
shown at the right. [1.3]
a

Law of Cosines
y The Law of Cosines can be used to solve triangles in which two sides and the included
B(a cos C, a sin C) angle (SAS) are known or in which three sides (SSS) are known. Consider the triangle in
Figure 7.11. The height BD is drawn from B perpendicular to the x-axis. The triangle BDA
is a right triangle, and the coordinates of B are 1a cos C, a sin C2. The coordinates of A are
a c 1b, 02. Using the distance formula, we can find the distance c.

c = 21a cos C - b22 + 1a sin C - 022

C
A(b, 0)
D (0, 0) b x
c2 = a2 cos2 C - 2ab cos C + b2 + a2 sin2 C
c2 = a21cos2 C + sin2 C2 + b2 - 2ab cos C
Figure 7.11

c2 = a2 + b2 - 2ab cos C

Law of Cosines
If A, B, and C are the measures of the angles of a triangle and a, b, and c are the
lengths of the sides opposite these angles, then

c2 = a2 + b2 - 2ab cos C
a2 = b2 + c2 - 2bc cos A
b2 = a2 + c2 - 2ac cos B
 7.2 LAW OF COSINES AND AREA 593

EXAMPLE 1 Use the Law of Cosines (SAS)

In triangle ABC, B = 110.0°, a = 10.0 centimeters, and c = 15.0 centimeters. See
Figure 7.12. Find b.
A
Solution
The Law of Cosines can be used because two sides and the included angle are known.
b2 = a2 + c2 - 2ac cos B
b
c = 15.0 cm = 10.02 + 15.02 - 2110.02115.02 cos 110.0°

110.0°
b = 210.02 + 15.02 - 2110.02115.02 cos 110.0°
b L 20.7 centimeters
B a = 10.0 cm C

Figure 7.12 Try Exercise 12, page 598

In the next example, we know the length of each side, but we do not know the measure
of any of the angles.

EXAMPLE 2 Use the Law of Cosines (SSS)

In triangle ABC, a = 32 feet, b = 20 feet, and c = 40 feet. Find B. This is the SSS case.

Solution

b2 = a2 + c2 - 2ac cos B
a2 + c2 - b2
cos B = • Solve for cos B.
2ac
322 + 402 - 202
= • Substitute for a, b, and c.
213221402
322 + 402 - 202
B = cos-1 a b • Solve for angle B.
213221402
B L 30° • To the nearest degree

Try Exercise 18, page 598

N
EXAMPLE 3 Solve an Application Using the Law of Cosines
N

B 138° A boat sailed 3.0 miles at a heading of 78° and then turned to a heading of 138° and
78° 3.0 mi
sailed another 4.3 miles. Find the distance and the bearing of the boat from the starting
A 78°
point.
α
4.3 mi
b
Solution
Sketch a diagram (see Figure 7.13). First find the measure of angle B in triangle ABC.
C B = 78° + 1180° - 138°2 = 120°
Figure 7.13 (continued)
594 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY

Use the Law of Cosines first to find b and then to find A.

b2 = a2 + c2 - 2ac cos B
= 4.32 + 3.02 - 214.3213.02 cos 120° • Substitute for a, c, and B.

b = 24.32 + 3.02 - 214.3213.02 cos 120°

b L 6.4 miles
b2 + c2 - a2
cos A = • Solve the Law of Cosines
2bc for cos A.
b2 + c2 - a2 6.42 + 3.02 - 4.32
A = cos-1 a b L cos-1 a b L 35°
2bc 12216.4213.02
Study tip
The bearing of the present position of the boat from the starting point A can be deter-
The measure of angle A in
Example 3 can also be mined by calculating the measure of angle a in Figure 7.13, on page 593.
determined by using the Law a L 180° - 178° + 35°2 = 67°
of Sines.
The distance is approximately 6.4 miles, and the bearing (to the nearest degree) is
S67°E.
Try Exercise 52, page 599

There are five different cases that we may encounter when solving an oblique triangle.
Each case is listed below under the law that can be used to solve the triangle.

Choosing Between the Law of Sines and the Law of Cosines

Apply the Law of Sines to solve an oblique triangle for each of the following cases.
ASA The measures of two angles of the triangle and the length of the
included side are known.
AAS The measures of two angles of the triangle and the length of a side
opposite one of these angles are known.
SSA The lengths of two sides of the triangle and the measure of an angle
opposite one of these sides are known. This case is called the ambigu-
ous case. It may yield one solution, two solutions, or no solution.
Apply the Law of Cosines to solve an oblique triangle for each of the following cases.
SSS The lengths of all three sides of the triangle are known. After find-
ing the measure of an angle, you can complete your solution by
using the Law of Sines.
SAS The lengths of two sides of the triangle and the measure of the
included angle are known. After finding the measure of the third
side, you can complete your solution by using the Law of Sines.

Question • In triangle ABC, A = 40°, C = 60°, and b = 114. Should you use the Law of Sines
or the Law of Cosines to solve this triangle?

Answer • Because the measure of two angles and the length of the included side are given, the
triangle can be solved by using the Law of Sines.
 7.2 LAW OF COSINES AND AREA 595

B
Area of a Triangle
1
a c The formula A = bh can be used to find the area of a triangle when the base and height
h 2
are given. In this section we will find the areas of triangles when the height is not given.
We will use K for the area of a triangle because the letter A is often used to represent the
C b A
measure of an angle.
Acute triangle Consider the areas of the acute and obtuse triangles in Figure 7.14.
B
Height of each triangle: h = c sin A
c 1
h a Area of each triangle: K = bh
2
C b A
1
Obtuse triangle
K = bc sin A • Substitute for h.
2
Figure 7.14 Thus we have established the following theorem.

Study tip Area of a Triangle

Because each formula requires
two sides and the included angle, The area K of triangle ABC is one-half the product of the lengths of any two sides
it is necessary to learn only one and the sine of the included angle. Thus
formula. 1 1 1
K = bc sin A K = ab sin C K = ac sin B
2 2 2

EXAMPLE 4 Find the Area of a Triangle

Given angle A = 62°, b = 12 meters, and c = 5.0 meters, find the area of triangle ABC.

Solution
In Figure 7.15, two sides and the included angle of the triangle are given. Using the
formula for area, we have
B

bc sin A = 112215.021sin 62°2 L 26 square meters

1 1
c = 5.0
K =
2 2
62° Try Exercise 30, page 598
A C
b = 12

Figure 7.15 When two angles and an included side are given, the Law of Sines is used to derive a
formula for the area of a triangle. First, solve for c in the Law of Sines.
c b
=
sin C sin B
b sin C
c =
sin B
1
Substitute for c in the formula K = bc sin A.
2

b sin A
1 1 b sin C
K = bc sin A = ba
2 2 sin B
b2 sin C sin A
K =
2 sin B
596 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY

In like manner, the following two alternative formulas can be derived for the area of a
triangle.

a2 sin B sin C c2 sin A sin B

K = and K =
2 sin A 2 sin C

EXAMPLE 5 Find the Area of a Triangle

Given A = 32°, C = 77°, and a = 14 inches, find the area of triangle ABC.

Solution
To use the preceding area formula, we need to know two angles and the included side.
Therefore, we need to determine the measure of angle B.
B = 180° - 32° - 77° = 71°
Thus
a2 sin B sin C 142 sin 71° sin 77°
K = = L 170 square inches
2 sin A 2 sin 32°
Try Exercise 32, page 598

Math Matters
Heron’s Formula
Recent findings indicate that
Heron’s formula for finding the The Law of Cosines can be used to derive Heron’s formula for the area of a triangle in
area of a triangle was first which three sides of the triangle are given.
discovered by Archimedes.
However, the formula is called
Heron’s formula in honor of the
geometer Heron of Alexandria
Heron’s Formula for Finding the Area of a Triangle
(A.D. 50), who gave an ingenious
proof of the theorem in his work If a, b, and c are the lengths of the sides of a triangle, then the area K of the
Metrica. Because Heron of triangle is
Alexandria was also known as
1a + b + c2
Hero, some texts refer to Heron’s 1
K = 2s1s - a21s - b21s - c2, where s =
formula as Hero’s formula. 2

Because s is one-half the perimeter of the triangle, it is called the semiperimeter.

EXAMPLE 6 Find an Area by Heron’s Formula

Find, to two significant digits, the area of the triangle with a = 7.0 meters,
b = 15 meters, and c = 12 meters.

Solution
Calculate the semiperimeter s.
a + b + c 7.0 + 15 + 12
s = = = 17
2 2
 7.2 LAW OF COSINES AND AREA 597

Use Heron’s formula.

K = 1s1s - a21s - b21s - c2
= 117117 - 7.02117 - 152117 - 122
= 11700 L 41 square meters

Try Exercise 40, page 598

EXAMPLE 7 Use Heron’s Formula to Solve an Application

The original portion of the Luxor Hotel in Las Vegas has the shape of a square
pyramid. Each face of the pyramid is an isosceles triangle with a base of 646 feet
and sides of length 576 feet. Assuming that the glass on the exterior of the Luxor Hotel
costs $35 per square foot, determine the cost of the glass, to the nearest $10,000, for
one of the triangular faces of the hotel.

Solution
The lengths (in feet) of the sides of a triangular face are a = 646, b = 576, and
c = 576.
a + b + c 646 + 576 + 576
s = = = 899 feet
2 2
K = 1s1s - a21s - b21s - c2

= 18991899 - 64621899 - 57621899 - 5762

= 123,729,318,063
L 154,043 square feet
Macduff Everton/CORBIS

The cost C of the glass is the product of the cost per square foot and the area.
C L 35 # 154,043 = 5,391,505
The approximate cost of the glass for one face of the Luxor Hotel is $5,390,000.
The pyramid portion of the Luxor Try Exercise 60, page 600
Hotel in Las Vegas, Nevada

EXERCISE SET 7.2

In Exercises 1 to 52, round answers according to the 5. b = 60, c = 84, A = 13°
rounding conventions on page 448.
6. a = 122, c = 144, B = 48°
In Exercises 1 to 14, find the third side of the triangle.
1. a = 12, b = 18, C = 44° 7. a = 9.0, b = 7.0, C = 72°

2. b = 30, c = 24, A = 120° 8. b = 12, c = 22, A = 55°

3. a = 120, c = 180, B = 56° 9. a = 4.6, b = 7.2, C = 124°

4. a = 400, b = 620, C = 116° 10. b = 12.3, c = 14.5, A = 6.5°

598 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY

11. a = 25.9, c = 33.4, B = 84.0° 34. a = 32, b = 24, c = 36

12. a = 14.2, b = 9.30, C = 9.20° 35. B = 54.3°, a = 22.4, b = 26.9

13. a = 122, c = 55.9, B = 44.2° 36. C = 18.2°, b = 13.4, a = 9.84

14. b = 444.8, c = 389.6, A = 78.44° 37. A = 116°, B = 34°, c = 8.5

38. B = 42.8°, C = 76.3°, c = 17.9

In Exercises 15 to 24, given three sides of a triangle,
find the specified angle. 39. a = 3.6, b = 4.2, c = 4.8

15. a = 25, b = 32, c = 40; find A.

40. a = 10.2, b = 13.3, c = 15.4

16. a = 60, b = 88, c = 120; find B.

41. Distance Between Airports A plane leaves airport A and
travels 560 miles to airport B at a bearing of N32°E. The plane
17. a = 8.0, b = 9.0, c = 12; find C.
leaves airport B and travels to airport C 320 miles away at a
bearing of S72°E. Find the distance from airport A to airport C.
18. a = 108, b = 132, c = 160; find A.
42. Length of a Street A developer owns a triangular lot at the
19. a = 80.0, b = 92.0, c = 124; find B.
intersection of two streets. The streets meet at an angle of
72°, and the lot has 300 feet of frontage along one street and
20. a = 166, b = 124, c = 139; find B.
416 feet of frontage along the other street. Find the length of
the third side of the lot.
21. a = 1025, b = 625.0, c = 1420; find C.
43. Baseball In a baseball game, a batter hits a ground ball 26 feet
22. a = 4.7, b = 3.2, c = 5.9; find A.
in the direction of the pitcher’s mound. See the figure below.
The pitcher runs forward and reaches for the ball. At that
23. a = 32.5, b = 40.1, c = 29.6; find B.
moment, how far is the ball from first base? (Note: A baseball
infield is a square that measures 90 feet on each side.)
24. a = 112.4, b = 96.80, c = 129.2; find C.

In Exercises 25 to 28, solve the triangle.

25. A = 39.4°, b = 15.5, c = 17.2

26. C = 98.4°, a = 141, b = 92.3

26 ft ft
90
27. a = 83.6, b = 144, c = 98.1

28. a = 25.4, b = 36.3, c = 38.2

44. B-2 Bomber The leading edge of each wing of the B-2
In Exercises 29 to 40, find the area of the given triangle. Stealth Bomber measures 105.6 feet in length. The angle
Round each area to the same number of significant digits between the wing’s leading edges 1 ∠ABC2 is 109.05°. What is
given for each of the given sides. the wing span (the distance from A to C) of the B-2 Bomber?

29. A = 105°, b = 12, c = 24 B

30. B = 127°, a = 32, c = 25

31. A = 42°, B = 76°, c = 12 A C

B-2 Stealth Bomber

32. B = 102°, C = 27°, a = 8.5
45. Angle Between the Diagonals of a Box The rectangular
33. a = 16, b = 12, c = 14 box in the following figure measures 6.50 feet by 3.25 feet
 7.2 LAW OF COSINES AND AREA 599

by 4.75 feet. Find the measure of the angle u that is formed 51. Distance to a Plane A plane traveling at 180 miles per hour
by the union of the diagonal shown on the front of the box passes 400 feet directly over an observer. The plane is traveling
and the diagonal shown on the right side of the box. along a straight path with an angle of elevation of 14°. Find the
distance of the plane from the observer 10 seconds after the
plane has passed directly overhead.

4.75 ft 52. Distance Between Ships A ship leaves a port at a speed of

16 miles per hour at a heading of 32°. One hour later another
ship leaves the port at a speed of 22 miles per hour at a heading
of 254°. Find the distance between the ships 4 hours after the
θ
first ship leaves the port.
3.25 ft

6.50 ft 53. Distance and Bearing from a Starting Point A plane flew
181 miles at a heading of 108.5° and then turned to a heading
46. Submarine Rescue Mission Use the distances shown in the of 124.6° and flew another 225 miles. Find the distance and the
following figure to determine the depth of the submarine bearing of the plane from the starting point.
below the surface of the water. Assume that the line segment
between the surface ships is directly above the submarine. 54. Engine Design An engine has a 16-centimeter connecting
rod that is attached to a rotating crank with a 4-centimeter
615 ft
radius. See the following figure.

499 ft 629 ft Piston

C
16 cm
4 cm
Submarine
c
B
A
47. Distance Between Ships Two ships left a port at the same
time. One ship traveled at a speed of 18 miles per hour at a
heading of 318°. The other ship traveled at a speed of 22 miles
per hour at a heading of 198°. Find the distance between the a. Use the Law of Cosines to find an equation that relates c
two ships after 10 hours of travel. and A.

b. Use the quadratic formula to solve the equation in

48. Distance Across a Lake Find the distance across the lake
a. for c.
shown in the figure.
c. Use the equation from b. to find c when A = 55°. Round to
A the nearest centimeter.

d. Use the Law of Sines to find c when A = 55°. Round to the

136 m nearest centimeter. How does this result compare with the
result in c.?

B 78.0° 55. Area of a Triangular Lot Find the area of a triangular piece
of land that is bounded by sides of 236 meters, 620 meters, and
814 meters. Round to the nearest hundred square meters.
162 m
56. Geometry Find the exact area of a parallelogram with sides of
exactly 8 feet and 12 feet. The shorter diagonal is exactly 10 feet.
C
57. Geometry Find the exact area of a square inscribed in a cir-
49. Geometry A regular hexagon is inscribed in a circle with a cle with a radius of exactly 9 inches.
radius of exactly 40 centimeters. Find the exact length of one
side of the hexagon. 58. Geometry Find the exact area of a regular hexagon inscribed
in a circle with a radius of exactly 24 centimeters.
50. Angle Between Boundaries of a Lot A triangular city lot
59. Cost of a Lot A commercial piece of real estate is priced at
has sides of 224 feet, 182 feet, and 165 feet. Find the angle
$2.20 per square foot. Find, to the nearest $1000, the cost of a
between the longer two sides of the lot.
triangular lot measuring 212 feet by 185 feet by 240 feet.
600 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY

60. Cost of a Lot An industrial piece of real estate is priced at Use Mollweide’s formula to determine whether either ABC
$4.15 per square foot. Find, to the nearest $1000, the cost of a or DEF has an incorrect dimension.
triangular lot measuring 324 feet by 516 feet by 412 feet.
64. Check Dimensions of Trusses The following diagram
61. Area of a Pasture Find the number of acres in a pasture shows some of the steel trusses in a railroad bridge.
whose shape is a triangle measuring 800 feet by 1020 feet by
F
680 feet. Round to the nearest hundredth of an acre. (An acre D
is 43,560 square feet.) C
A

62. Area of a Housing Tract Find the number of acres in a

housing tract whose shape is a triangle measuring 420 yards by B E
540 yards by 500 yards. Round to the nearest tenth of an acre.
(An acre is 4840 square yards.) A structural engineer has determined the following dimensions
for ABC and DEF:
The identity at the right is one ABC: A = 34.1°, B = 66.2°, C = 79.7°
b
A B
of Mollweide’s formulas. sina a = 9.23 feet, b = 15.1 feet, c = 16.2 feet
a b 2
It applies to any triangle ABC. !
c DEF: D = 45.0°, E = 56.2°, F = 78.8°
cos a b
C
2 d = 13.6 feet, e = 16.0 feet, f = 18.9 feet
The formula is intriguing because it contains the Use Mollweide’s formula to determine whether either ABC
dimensions of all angles and all sides of triangle ABC. or DEF has an incorrect dimension.
The formula can be used to check whether a triangle has
been solved correctly. Substitute the dimensions of a 65. Find the measure of the angle formed by the sides P1P2 and
given triangle into the formula and compare the value of P1P3 of a triangle with vertices at P11 -2, 42, P212, 12, and
the left side of the formula with the value of the right P314, - 32.
side. If the two results are not reasonably close, then you
know that at least one dimension is incorrect. The results 66. Given a triangle ABC, prove that
generally will not be identical because each dimension
a2 = b2 + c2 - 2bc cos A
is an approximation. In Exercises 63 and 64, you can
assume that a triangle has an incorrect dimension if the 67. Use the Law of Cosines to show that for any triangle ABC,
value of the left side and the value of the right side of 1b + c - a21b + c + a2
the above formula differ by more than 0.02. cos A = - 1
2bc
63. Check Dimensions of Trusses The following diagram 68. Prove that K = xy sin A for a parallelogram, where x and y are
shows some of the steel trusses in an airplane hangar. the lengths of adjacent sides, A is the measure of the angle
F between side x and side y, and K is the area of the parallelogram.
D
C
A
69. Find the volume of
the triangular prism
B E
shown in the
figure.
72° 4 in.
4 in. 18 in.

70. Show that the area of the circumscribed triangle in the fol-
a + b + c
lowing figure is K = rs, where s = .
2
B
An architect has determined the following dimensions for
ABC and DEF:
c
ABC: A = 53.5°, B = 86.5°, C = 40.0° a
a = 13.0 feet, b = 16.1 feet, c = 10.4 feet
r
DEF: D = 52.1°, E = 59.9°, F = 68.0°
d = 17.2 feet, e = 21.3 feet, f = 22.8 feet A b C