Download Law of Cosines and Area

592
CHAPTER 7
APPLICATIONS OF TRIGONOMETRY
SECTION 7.2
Law of Cosines and Area
Law of Cosines
Area of a Triangle
Heron’s Formula
PREPARE FOR THIS SECTION
Prepare for this section by completing the following exercises. The answers can be found
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
. [P.2]
2
PS6. State a relationship between the lengths a, b, and c in the triangle
shown at the right. [1.3]
s =
c
b
a
Law of Cosines
y
The Law of Cosines can be used to solve triangles in which two sides and the included
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
1b, 02. Using the distance formula, we can find the distance c.
B(a cos C, a sin C)
c
a
C
D
(0, 0)
c = 21a cos C - b22 + 1a sin C - 022
A(b, 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
EXAMPLE 1
LAW OF COSINES AND AREA
593
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°
b = 210.02 + 15.02 - 2110.02115.02 cos 110.0°
b L 20.7 centimeters
110.0°
B
a = 10.0 cm
C
Try Exercise 12, page 598
Figure 7.12
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 =
2ac
=
322 + 402 - 202
213221402
B = cos-1 a
322 + 402 - 202
b
213221402
B L 30°
• Solve for cos B.
• Substitute for a, b, and c.
• Solve for angle B.
• To the nearest degree
Try Exercise 18, page 598
N
EXAMPLE 3
N
78° 3.0 mi
B
A boat sailed 3.0 miles at a heading of 78° and then turned to a heading of 138° and
sailed another 4.3 miles. Find the distance and the bearing of the boat from the starting
point.
138°
78°
A
α
Solve an Application Using the Law of Cosines
4.3 mi
Solution
Sketch a diagram (see Figure 7.13). First find the measure of angle B in triangle ABC.
b
C
Figure 7.13
B = 78° + 1180° - 138°2 = 120°
(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
cos A =
b2 + c2 - a2
2bc
A = cos-1 a
Study tip
The measure of angle A in
Example 3 can also be
determined by using the Law
of Sines.
• Solve the Law of Cosines
for cos A.
b2 + c2 - a2
6.42 + 3.02 - 4.32
b L cos-1 a
b L 35°
2bc
12216.4213.02
The bearing of the present position of the boat from the starting point A can be determined by calculating the measure of angle a in Figure 7.13, on page 593.
a L 180° - 178° + 35°2 = 67°
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 ambiguous 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 finding 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
1
bh can be used to find the area of a triangle when the base and height
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
measure of an angle.
Consider the areas of the acute and obtuse triangles in Figure 7.14.
The formula A =
c
h
C
b
A
Acute triangle
B
Height of each triangle:
c
h
595
Area of a Triangle
B
a
LAW OF COSINES AND AREA
1
bh
2
1
K = bc sin A
2
Thus we have established the following theorem.
Area of each triangle:
a
C
b
h = c sin A
A
Obtuse triangle
Figure 7.14
Study tip
K =
• Substitute for h.
Area of a Triangle
Because each formula requires
two sides and the included angle,
it is necessary to learn only one
formula.
The area K of triangle ABC is one-half the product of the lengths of any two sides
and the sine of the included angle. Thus
1
1
1
K = ac sin B
K = bc sin A
K = ab sin C
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
K =
c = 5.0
62°
A
b = 12
Figure 7.15
C
1
1
bc sin A = 112215.021sin 62°2 L 26 square meters
2
2
Try Exercise 30, page 598
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
c =
Substitute for c in the formula K =
b sin C
sin B
1
bc sin A.
2
K =
1 b sin C
1
bc sin A = ba
b sin A
2
2
sin B
K =
b2 sin C sin A
2 sin B
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CHAPTER 7
APPLICATIONS OF TRIGONOMETRY
In like manner, the following two alternative formulas can be derived for the area of a
triangle.
K =
EXAMPLE 5
a2 sin B sin C
2 sin A
and
K =
c2 sin A sin B
2 sin C
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
K =
142 sin 71° sin 77°
a2 sin B sin C
=
L 170 square inches
2 sin A
2 sin 32°
Try Exercise 32, page 598
Math Matters
Recent findings indicate that
Heron’s formula for finding the
area of a triangle was first
discovered by Archimedes.
However, the formula is called
Heron’s formula in honor of the
geometer Heron of Alexandria
(A.D. 50), who gave an ingenious
proof of the theorem in his work
Metrica. Because Heron of
Alexandria was also known as
Hero, some texts refer to Heron’s
formula as Hero’s formula.
Heron’s Formula
The Law of Cosines can be used to derive Heron’s formula for the area of a triangle in
which three sides of the triangle are given.
Heron’s Formula for Finding the Area of a Triangle
If a, b, and c are the lengths of the sides of a triangle, then the area K of the
triangle is
K = 2s1s - a21s - b21s - c2,
where s =
1
1a + b + c2
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.
s =
7.0 + 15 + 12
a + b + c
=
= 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.
s =
a + b + c
646 + 576 + 576
=
= 899 feet
2
2
K = 1s1s - a21s - b21s - c2
Macduff Everton/CORBIS
= 18991899 - 64621899 - 57621899 - 5762
The pyramid portion of the Luxor
Hotel in Las Vegas, Nevada
= 123,729,318,063
L 154,043 square feet
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.
Try Exercise 60, page 600
EXERCISE SET 7.2
In Exercises 1 to 52, round answers according to the
rounding conventions on page 448.
In Exercises 1 to 14, find the third side of the triangle.
5. b = 60, c = 84, A = 13°
6. a = 122, c = 144, B = 48°
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°
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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.
15. a = 25, b = 32, c = 40; find A.
39. a = 3.6, b = 4.2, c = 4.8
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
17. a = 8.0, b = 9.0, c = 12; find C.
travels 560 miles to airport B at a bearing of N32°E. The plane
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.
19. a = 80.0, b = 92.0, c = 124; find B.
42. Length of a Street A developer owns a triangular lot at the
intersection of two streets. The streets meet at an angle of
72°, and the lot has 300 feet of frontage along one street and
416 feet of frontage along the other street. Find the length of
the third side of the lot.
20. a = 166, b = 124, c = 139; find B.
21. a = 1025, b = 625.0, c = 1420; find C.
22. a = 4.7, b = 3.2, c = 5.9; find A.
43. Baseball In a baseball game, a batter hits a ground ball 26 feet
in the direction of the pitcher’s mound. See the figure below.
The pitcher runs forward and reaches for the ball. At that
moment, how far is the ball from first base? (Note: A baseball
infield is a square that measures 90 feet on each side.)
23. a = 32.5, b = 40.1, c = 29.6; find B.
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
27. a = 83.6, b = 144, c = 98.1
90
ft
28. a = 25.4, b = 36.3, c = 38.2
44.
In Exercises 29 to 40, find the area of the given triangle.
Round each area to the same number of significant digits
given for each of the given sides.
B-2 Bomber The leading edge of each wing of the B-2
Stealth Bomber measures 105.6 feet in length. The angle
between the wing’s leading edges 1 ∠ABC2 is 109.05°. What is
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
by 4.75 feet. Find the measure of the angle u that is formed
by the union of the diagonal shown on the front of the box
and the diagonal shown on the right side of the box.
4.75 ft
LAW OF COSINES AND AREA
599
51. Distance to a Plane A plane traveling at 180 miles per hour
passes 400 feet directly over an observer. The plane is traveling
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.
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
first ship leaves the port.
θ
3.25 ft
53. Distance and Bearing from a Starting Point A plane flew
6.50 ft
46. Submarine Rescue Mission Use the distances shown in the
following figure to determine the depth of the submarine
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
radius. See the following figure.
615 ft
499 ft
181 miles at a heading of 108.5° and then turned to a heading
of 124.6° and flew another 225 miles. Find the distance and the
bearing of the plane from the starting point.
Piston
629 ft
C
16 cm
4 cm
Submarine
c
A
B
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
two ships after 10 hours of travel.
48. Distance Across a Lake Find the distance across the lake
shown in the figure.
a. Use the Law of Cosines to find an equation that relates c
and A.
b. Use the quadratic formula to solve the equation in
a. for c.
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
B
nearest centimeter. How does this result compare with the
result in c.?
55. Area of a Triangular Lot Find the area of a triangular piece
78.0°
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 cir49. Geometry A regular hexagon is inscribed in a circle with a
radius of exactly 40 centimeters. Find the exact length of one
side of the hexagon.
50. Angle Between Boundaries of a Lot A triangular city lot
has sides of 224 feet, 182 feet, and 165 feet. Find the angle
between the longer two sides of the lot.
cle with a radius of exactly 9 inches.
58. Geometry Find the exact area of a regular hexagon inscribed
in a circle with a radius of exactly 24 centimeters.
59. Cost of a Lot A commercial piece of real estate is priced at
$2.20 per square foot. Find, to the nearest $1000, the cost of a
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
or DEF has an incorrect dimension.
$4.15 per square foot. Find, to the nearest $1000, the cost of a
triangular lot measuring 324 feet by 516 feet by 412 feet.
64.
61. Area of a Pasture Find the number of acres in a pasture
ABC
Check Dimensions of Trusses The following diagram
shows some of the steel trusses in a railroad bridge.
whose shape is a triangle measuring 800 feet by 1020 feet by
680 feet. Round to the nearest hundredth of an acre. (An acre
is 43,560 square feet.)
F
D
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
540 yards by 500 yards. Round to the nearest tenth of an acre.
(An acre is 4840 square yards.)
The identity at the right is one
of Mollweide’s formulas.
It applies to any triangle ABC.
a
sina
b
A
B
2
b
!
c
C
cos a b
2
The formula is intriguing because it contains the
dimensions of all angles and all sides of triangle ABC.
The formula can be used to check whether a triangle has
been solved correctly. Substitute the dimensions of a
given triangle into the formula and compare the value of
the left side of the formula with the value of the right
side. If the two results are not reasonably close, then you
know that at least one dimension is incorrect. The results
generally will not be identical because each dimension
is an approximation. In Exercises 63 and 64, you can
assume that a triangle has an incorrect dimension if the
value of the left side and the value of the right side of
the above formula differ by more than 0.02.
63.
Check Dimensions of Trusses The following diagram
shows some of the steel trusses in an airplane hangar.
C
A
F
D
E
B
A structural engineer has determined the following dimensions
for ABC and DEF:
ABC:
A = 34.1°, B = 66.2°, C = 79.7°
a = 9.23 feet, b = 15.1 feet, c = 16.2 feet
DEF:
D = 45.0°, E = 56.2°, F = 78.8°
d = 13.6 feet, e = 16.0 feet, f = 18.9 feet
Use Mollweide’s formula to determine whether either
or DEF has an incorrect dimension.
ABC
65. Find the measure of the angle formed by the sides P1P2 and
P1P3 of a triangle with vertices at P11 -2, 42, P212, 12, and
P314, - 32.
66. Given a triangle ABC, prove that
a2 = b2 + c2 - 2bc cos A
67. Use the Law of Cosines to show that for any triangle ABC,
cos A =
1b + c - a21b + c + a2
2bc
- 1
68. Prove that K = xy sin A for a parallelogram, where x and y are
the lengths of adjacent sides, A is the measure of the angle
between side x and side y, and K is the area of the parallelogram.
69. Find the volume of
B
E
the triangular prism
shown in the
figure.
72°
4 in.
4 in.
18 in.
70. Show that the area of the circumscribed triangle in the fol-
lowing figure is K = rs, where s =
a + b + c
.
2
B
An architect has determined the following dimensions for
ABC and DEF:
ABC:
c
A = 53.5°, B = 86.5°, C = 40.0°
a = 13.0 feet, b = 16.1 feet, c = 10.4 feet
DEF: D = 52.1°, E = 59.9°, F = 68.0°
d = 17.2 feet, e = 21.3 feet, f = 22.8 feet
a
r
A
b
C