4-7 The Law of Sines and the Law of Cosines
Therefore,
,f
55.5, and g
64.5.
Solve each triangle. Round to the nearest
tenth, if necessary.
3. 1. SOLUTION: SOLUTION: Because two angles are given, K = 180 – (40 +
58 ) or 82 . Use the Law of Sines to find j and ℓ.
Because two angles are given, A = 180 – (110 +
38 ) or 32 . Use the Law of Sines to find a and b.
Therefore,
,a
11.2, and b
Therefore,
19.8.
2. ,j
16.2, and 21.4.
4. SOLUTION: SOLUTION: Because two angles are given, H = 180 – (53 +
112 ) or 15 . Use the Law of Sines to find f and g.
Therefore,
,f
55.5, and g
Because two angles are given, S = 180 – (62 +
56 ) or 62 . Use the Law of Sines to find r and s.
64.5.
Therefore,
,r
6.6, and s
7.
5. SOLUTION: 3. SOLUTION: Because two angles are given, T = 180 – (12 +
148 ) or 20 . Use the Law of Sines to find t and u.
Because two angles are given, K = 180 – (40 +
58 ) or 82 . Use the Law of Sines to find j and ℓ.
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Page 1
4-7 The Law of Sines and the Law of Cosines
Therefore,
,r
6.6, and s
7.
Therefore,
b
12.8, and c
28.7.
7. GOLF A golfer misses a 12-foot putt by putting 3º
off course. The hole now lies at a 129º angle
between the ball and its spot before the putt. What
distance does the golfer need to putt in order to
make the shot?
5. SOLUTION: Because two angles are given, T = 180 – (12 +
148 ) or 20 . Use the Law of Sines to find t and u.
SOLUTION: Draw a diagram of a triangle with two angle
measures of 3 and 129 and an included side
length of 12 feet.
Therefore,
t
29.0, and u
17.7.
(not drawn to scale)
Because two angles are given, the missing angle is
180° − (3° + 129°) or 48°. Use the Law of Sines to find x.
6. SOLUTION: Because two angles are given, D = 180 – (25 +
72 ) or 83º. Use the Law of Sines to find band c.
Therefore, the distance the golfer needs to putt is
about 0.85 ft.
8. ARCHITECTURE An architect’s client wants to
Therefore,
b
12.8, and c
28.7.
7. GOLF A golfer misses a 12-foot putt by putting 3º
off course. The hole now lies at a 129º angle
between the ball and its spot before the putt. What
distance does the golfer need to putt in order to
make the shot?
build a home based on the architect Jon Lautner’s
Sheats-Goldstein House. The length of the patio will
be 60 feet. The left side of the roof will be at a 49º
angle of elevation, and the right side will be at an 18º
angle of elevation. Determine the lengths of the left
and right sides of the roof and the angle at which
they will meet.
SOLUTION: Draw a diagram of a triangle with two angle
measures of 3 and 129 and an included side
length of 12 feet.
SOLUTION: Draw a diagram of a triangle with two angle
measures of 49 and 18 and an included side
length of 60 feet.
drawn
eSolutions Manual - Powered by (not
Cognero
to scale)
Page 2
Because two angles are given, the missing angle is
b. Determine the distance of a direct flight to the
destination
golfer
to putt
is
4-7 Therefore,
The Lawtheofdistance
Sinesthe
and
theneeds
Law
of Cosines
about 0.85 ft.
SOLUTION: a. Draw a diagram to model the situation.
8. ARCHITECTURE An architect’s client wants to
build a home based on the architect Jon Lautner’s
Sheats-Goldstein House. The length of the patio will
be 60 feet. The left side of the roof will be at a 49º
angle of elevation, and the right side will be at an 18º
angle of elevation. Determine the lengths of the left
and right sides of the roof and the angle at which
they will meet.
(Not drawn to scale)
Because two angles are given, the missing angle is
180° − (157° + 8°) or 15°. Use the Law of Sines to find x.
SOLUTION: Draw a diagram of a triangle with two angle
measures of 49 and 18 and an included side
length of 60 feet.
Because two angles are given, θ is 180° − (49° +
18°) or 113°. Use the Law of Sines to find x and y.
Therefore, the distance of the flight is 90 + 48.4 or
about 138.4 miles.
b. Find the length of the side opposite the 157
angle.
Therefore, the left and right sides of the roof are
about 20.1 and 49.2 feet, respectively, and the angle
at which they meet is about 113 .
9. TRAVEL For the initial 90 miles of a flight, the pilot
heads 8º off course in order to avoid a storm. The
pilot then changes direction to head toward the
destination for the remainder of the flight, making a
157 angle to the first flight course.
a. Determine the total distance of the flight.
b. Determine the distance of a direct flight to the
destination
SOLUTION: a. Draw a diagram to model the situation.
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Therefore, the distance of a direct flight is about
135.9 miles.
Find all solutions for the given triangle, if
possible. If no solution exists, write no solution.
Round side lengths to the nearest tenth and
angle measures to the nearest degree.
10. a = 9, b = 7, A = 108
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Page 3
a direct
is about
4-7 Therefore,
The Lawtheofdistance
Sinesofand
theflight
Law
of Cosines
135.9 miles.
Find all solutions for the given triangle, if
possible. If no solution exists, write no solution.
Round side lengths to the nearest tenth and
angle measures to the nearest degree.
10. a = 9, b = 7, A = 108
SOLUTION: Therefore, the remaining measures of
B 48 , C 24 , and c 3.9.
are
11. a = 14, b = 15, A = 117
SOLUTION: A is obtuse and a < b because 14 <15. Therefore,
this problem has no solution.
12. a = 18, b = 12, A = 27
Draw a diagram of a triangle with the given
dimensions.
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
(Not drawn to scale)
Notice that A is obtuse and a > b because 9 > 7.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Notice that A is acute and a > b because 18 > 12.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Because two angles are now known, C ≈ 180 –
(108 + 48 ) or about 24 . Apply the Law of Sines
to find c.
Because two angles are now known, C 180º –
(27º + 17.62º) or about 135.38º. Apply the Law of
Sines to find c.
Therefore, the remaining measures of
B 48 , C 24 , and c 3.9.
are
11. a = 14, b = 15, A = 117
SOLUTION: A is obtuse and a < b because 14 <15. Therefore,
this problem has no solution.
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12. a = 18, b = 12, A = 27
SOLUTION: Therefore, the remaining measures of ABC are
B 18 , C 135 , and c 27.8.
13. a = 35, b = 24, A = 92
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Page 4
measures
ABC
4-7 Therefore,
The Lawtheofremaining
Sines and
theofLaw
ofare
Cosines
B
18 , C
135 , and c
27.8.
13. a = 35, b = 24, A = 92
Therefore, the remaining measures of
B 43 , C 45 , and c 24.7.
are
14. a = 14, b = 6, A = 145
SOLUTION: SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Draw a diagram of a triangle with the given
dimensions.
Notice that A is obtuse and a > b because 35 > 24.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Notice that A is obtuse and a > b because 14 > 6.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Because two angles are now known, C 180 –
(92 + 43.26 ) or about 44.74 . Apply the Law of
Sines to find c.
Because two angles are now known, C ≈ 180 –
(145 + 14.23 ) or about 20.77 . Apply the Law
of Sines to find c.
Therefore, the remaining measures of
B 43 , C 45 , and c 24.7.
14. a = 14, b = 6, A = 145
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
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are
Therefore, the remaining measures of
B 14 , C 21 , and c 8.7.
are
15. a = 19, b = 38, A = 30
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Page 5
measures
4-7 Therefore,
The Lawtheofremaining
Sines and
theofLaw ofare
Cosines
B
14 , C
21 , and c
8.7.
15. a = 19, b = 38, A = 30
Therefore, the remaining measures of
B = 90 , C = 30 , and c 32.9.
are
16. a = 5, b = 6, A = 63
SOLUTION: SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Notice that A is acute and a < b because 5 < 6. So,
this problem has no solution, one solution, or two
solutions. Find h.
Notice that A is acute and b > a because 38 > 19.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Because a < h, no triangle can be formed with sides
a = 5, b = 6, and A = 63 . Therefore, this problem
has no solution.
17. a = 10, b =
, A = 45
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Because two angles are now known, C =180 –
(30 + 90 ) or 60 . Apply the Law of Sines to find
c.
Notice that A is acute and b > a because
> 10. Therefore, one solution exists. Apply the Law of
Sines to find B.
Therefore, the remaining measures of
B = 90 , C = 30 , and c 32.9.
are
16. a = 5, b = 6, A = 63
SOLUTION: Notice that A is acute and a < b because 5 < 6. So,
this problem has no solution, one solution, or two
solutions. Find h.
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Because a < h, no triangle can be formed with sides
Because two angles are now known, C ≈ 180 –
(45 + 90 ) or 45 . Apply the Law of Sines to find
c.
Page 6
4-7
Because a < h, no triangle can be formed with sides
aThe
= 5, Law
b = 6, and
A = 63 and
. Therefore,
this problem
of Sines
the Law
of Cosines
has no solution.
17. a = 10, b =
Therefore, the remaining measures of
B =90 , C = 45 , and c =10.
are
18. SKIING A ski lift rises at a 28 angle during the
, A = 45
first 20 feet up a mountain to achieve a height of 25
feet, which is the height maintained during the
remainder of the ride up the mountain. Determine
the length of cable needed for this initial rise.
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
SOLUTION: Notice that A is acute and b > a because
> 10. Therefore, one solution exists. Apply the Law of
Sines to find B.
In this problem, A = 28 , a = 25 ft, and b = 20 ft. So,
A is acute and a > b. Therefore, one solution exists.
Apply the Law of Sines to find B.
Because two angles are now known, C ≈ 180 –
(45 + 90 ) or 45 . Apply the Law of Sines to find
c.
Therefore, the remaining measures of
B =90 , C = 45 , and c =10.
are
18. SKIING A ski lift rises at a 28 angle during the
first 20 feet up a mountain to achieve a height of 25
feet, which is the height maintained during the
remainder of the ride up the mountain. Determine
the length of cable needed for this initial rise.
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Because two angles are now known, the angle
opposite x is 180 – (28 + 22.06 ) or about
129.94 . Apply the Law of Sines to find x.
Therefore, the length of cable needed for the initial
rise is about 41 feet.
Find two triangles with the given angle
measure and side lengths. Round side lengths
to the nearest tenth and angle measures to the
nearest degree.
19. A = 39 , a = 12, b = 17
SOLUTION: A is acute, and h = 17 sin 39 or about 10.7. Notice
that a < b because 12 < 17, and a > h because 12 >
10.7. Therefore, two different triangles are possible
Page 7
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
nearest degree.
19. A = 39 , a = 12, b = 17
SOLUTION: 4-7 The Law of Sines and the Law of Cosines
A is acute, and h = 17 sin 39 or about 10.7. Notice
that a < b because 12 < 17, and a > h because 12 >
10.7. Therefore, two different triangles are possible
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
Solution 2
B is obtuse.
Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Note that
. To find B', find an
obtuse angle with a sine that is also 0.8915. To do
this, subtract the measure given by your calculator to
nearest degree, 63º, from 180º. Therefore, B' is
approximately 117º.
Find C.
Solution 1
B is acute.
Apply the Law of Sines to find c.
Find B.
Therefore, the missing measures for acute
are
B 63 , C 78 , and c 18.7, while the missing
measures for obtuse
are B' 117 , C
24 , and c 7.8.
20. A = 26 , a = 5, b = 9
Find C.
SOLUTION: A is acute, and h =9 sin 26 or about 3.9. Notice
that a < b because 5 < 9, and a > h because 5 > 3.9.
Therefore, two different triangles are possible with
the given angle and side measures. Angle B will be
acute, and angle B' will be obtuse, as shown below.
Apply the Law of Sines to find c.
Solution 2
B is obtuse.
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Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Solution 1
Page 8
B is acute.
4-7
Make a reasonable sketch of each triangle and apply
the
Law
of Sines
to find each
Startof
with
the
The
Law
of Sines
andsolution.
the Law
Cosines
case in which B is acute.
Solution 1
nearest degree, 52 , from 180 . Therefore, B' is
approximately 128 .
Find C.
B is acute.
Apply the Law of Sines to find c.
Find B.
Therefore, the missing measures for acute
are B 52 , C 102 , and c 11.2, while the missing measures for obtuse
are B'
128 , C 26 , and c 5.0.
21. A = 61 , a = 14, b = 15
SOLUTION: A is acute, and h = 14sin 61 or about 12.2. Notice
that a < b because 14 < 15, and a > h because 14 >
12.2. Therefore, two different triangles are possible
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
Find C.
Apply the Law of Sines to find c.
Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Solution 2
B is obtuse.
Solution 1
B is acute.
Find B.
Note that
. To find B', find an
obtuse angle with a sine that is also 0.2630. To do
this, subtract the measure given by your calculator to
nearest degree, 52 , from 180 . Therefore, B' is
approximately 128 .
Find C.
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eSolutions
Page 9
Apply the Law of Sines to find c.
Find C.
4-7 The Law of Sines and the Law of Cosines
Find B.
Apply the Law of Sines to find c.
Therefore, the missing measures for acute
are
B 70 , C 49 , and c 12.2, while the missing
measures for obtuse
are B' 110 , C
9 , and c 2.5.
Find C.
Apply the Law of Sines to find c.
22. A = 47 , a = 25, b = 34
SOLUTION: A is acute, and h = 34 sin 47 or about 24.9. Notice
that a < b because 25 < 34, and a > h because 25 >
24.9. Therefore, two different triangles are possible
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
Solution 2
B is obtuse.
Note that
. To find B', find an
obtuse angle with a sine that is also 0.9371. To do
this, subtract the measure given by your calculator to
nearest degree, 70 , from 180 . Therefore, B' is
approximately 110 .
Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Solution 1
B is acute.
Find C.
Apply the Law of Sines to find c.
Find B.
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Therefore, the missing measures for acute
are
Page 10
Find C.
4-7 The Law of Sines and the Law of Cosines
Find B.
Apply the Law of Sines to find c.
Therefore, the missing measures for acute
are B 84 , C 49 , and c 25.8, while the missing measures for obtuse
are B'
96 , C 37 , and c 20.6.
Find C.
23. A = 54 , a = 31, b = 36
Apply the Law of Sines to find c.
SOLUTION: A is acute, and h = 36 sin 54º or about 29.1. Notice
that a < b because 31 < 36, and a > h because 31 >
29.1. Therefore, two different triangles are possible
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
Solution 2
B is obtuse.
Note that
. To find B', find an
obtuse angle with a sine that is also 0.9946. To do
this, subtract the measure given by your calculator to
nearest degree, 84º, from 180º. Therefore, B' is
approximately 96 .
Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Solution 1
B is acute.
Find C.
Find B.
Apply the Law of Sines to find c.
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Therefore, the missing measures for acute
Page 11
approximately 180º – 70º or 110º.
Find C.
4-7 The Law of Sines and the Law of Cosines
Find B.
Apply the Law of Sines to find c.
Therefore, the missing measures for acute
are B 70 , C 56 , and c 31.8, while the missing measures for obtuse
are B'
110 , C 16 , and c 10.6
Find C.
24. A = 18 , a = 8, b = 13
Apply the Law of Sines to find c.
SOLUTION: A is acute, and h = 13 sin 18 or about 4.01. Notice
that a < b because 8 < 13, and a > h because 8 >
4.01. Therefore, two different triangles are possible
with the given angle and side measures. Angle B will
be acute, and angle B' will be obtuse, as shown
below.
Solution 2
B is obtuse.
Make a reasonable sketch of each triangle and apply
the Law of Sines to find each solution. Start with the
case in which B is acute.
Note that
. To find B', find an
obtuse angle with a sine that is also 0.9395. To do
this, subtract the measure given by your calculator to
nearest degree, 70º, from 180º. Therefore, B' is
approximately 180º – 70º or 110º.
Find C.
Apply the Law of Sines to find c.
Solution 1
Find B.
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Therefore, the missing measures for acute
are B 70 , C 56 , and c 31.8, while B is acute.
Page 12
Find C.
4-7 The Law of Sines and the Law of Cosines
Find B.
Apply the Law of Sines to find c.
Therefore, the missing measures for acute
are B 30 , C 132 , and c 19.2, while the missing measures for obtuse
are B'
150 , C 12 , and c 5.4.
Find C.
25. BROADCASTING A radio tower located 38 Apply the Law of Sines to find c.
miles along Industrial Parkway transmits radio
broadcasts over a 30-mile radius. Industrial Parkway
intersects the interstate at a 41º angle. How far
along the interstate can vehicles pick up the
broadcasting signal?
Solution 2
B is obtuse.
SOLUTION: Note that
. To find B', find an
obtuse angle with a sine that is also 0.5022. To do
this, subtract the measure given by your calculator to
nearest degree, 30º, from 180º. Therefore, B' is
approximately 180º – 30º or 150º.
Draw a diagram of the situation where A represents
the location of the radio tower, B represents the
leftmost point on the interstate that is within the 30mile broadcasting radius, and C' represents the
rightmost point on the interstate within the
broadcasting radius.
Find C.
Apply the Law of Sines to find c.
Due to the information given, use the Law of Sines
to solve for B in ABC.
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Therefore, the missing measures for acute
Page 13
the distance between the two boats when the second
boat enters the radius of the lighthouse light.
4-7 Due
ThetoLaw
of Sines and the Law of Cosines
the information given, use the Law of Sines
to solve for B in ABC.
SOLUTION: Draw a diagram to represent the situation.
Because two angles are now known, A 180 –
(41 + 56.2 ) or about 82.8 . Apply the Law of
Sines again to find BC.
Due to the information given, use the Law of Sines
to solve for B in ABC.
Notice that ABC' is an isosceles triangle, so B ≈ C',
B ≈ 56.2º, and thus C' ≈ 56.2º. Because two angles
in ABC' are now known, A ≈ 180º – (56.2º + 56.2º) ≈
67.6º. Use the Law of Cosines to find BC'.
So, A ≈ 180º – (44º + 74.75º) or about 61.25º.
Notice that ABC is an isosceles triangle. Draw an
altitude from vertex A to BC'.
Therefore, vehicles can pick up the broadcasting
signal for about 33.4 miles along the interstate.
Use the sine function to find x.
26. BOATING The light from a lighthouse can be seen from an 18-mile radius. A boat is anchored so
that it can just see the light from the lighthouse. A
second boat is located 25 miles away from the
lighthouse and is headed straight toward it, making a
44º angle with the lighthouse and the first boat. Find
the distance between the two boats when the second
boat enters the radius of the lighthouse light.
SOLUTION: Draw a diagram to represent the situation.
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Therefore, the distance between the two boats when
the second boat enters the radius of the lighthouse
light is 2(9.17) or about 18.3 miles.
Solve each triangle. Round side lengths to the
nearest tenth and angle measures to the
nearest degree.
27. ABC, if A = 42 , b = 12, and c = 19
SOLUTION: Page 14
Use the Law of Cosines to find the missing side
measure.
Find the measure of the remaining angle.
4-7
Therefore, the distance between the two boats when
the
second
radius
of the
lighthouse
The
Lawboat
ofenters
Sinestheand
the
Law
of Cosines
light is 2(9.17) or about 18.3 miles.
Solve each triangle. Round side lengths to the
nearest tenth and angle measures to the
nearest degree.
27. ABC, if A = 42 , b = 12, and c = 19
SOLUTION: Therefore, B
39 , C
99 and a
12.9.
28. XYZ, if x = 5, y = 18, and z = 14
SOLUTION: Use the Law of Cosines to find a missing angle
measure.
Use the Law of Cosines to find the missing side
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Find the measure of the remaining angle.
Find the measure of the remaining angle.
Therefore, B
39 , C
99 and a
12.9.
28. XYZ, if x = 5, y = 18, and z = 14
SOLUTION: Use the Law of Cosines to find a missing angle
measure.
Therefore, X
11 , Y
137 and Z
32 .
29. PQR, if P = 73 , q = 7, and r = 15
SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
eSolutions Manual - Powered by Cognero
Page 15
Find the measure of the remaining angle.
Find the measure of the remaining angle.
4-7 The Law of Sines and the Law of Cosines
Therefore, X
11 , Y
137 and Z
32 .
Therefore, Q
27 , R
80 and p
14.6.
30. JKL, if J = 125 , k = 24, and l = 33
29. PQR, if P = 73 , q = 7, and r = 15
SOLUTION: SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Cosines to find the missing side
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Find the measure of the remaining angle.
Find the measure of the remaining angle.
Therefore, Q
27 , R
80 and p
14.6.
30. JKL, if J = 125 , k = 24, and l = 33
Therefore, K
23 , L
32 and j
50.7.
31. RST, if r = 35, s = 22, and t = 25
SOLUTION: SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Cosines to find the missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
eSolutions Manual - Powered by Cognero
Page 16
Find the measure of the remaining angle.
4-7 The Law of Sines and the Law of Cosines
Therefore, K
23 , L
32 and j
50.7.
31. RST, if r = 35, s = 22, and t = 25
Find the measure of the remaining angle.
Therefore, R
96 , S
39 and T
45 .
32. FGH, if f = 39, g = 50, and h = 64
SOLUTION: SOLUTION: Use the Law of Cosines to find the missing angle
measure.
Use the Law of Cosines to find the missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Find the measure of the remaining angle.
Therefore, R
96 , S
39 and T
45 .
32. FGH, if f = 39, g = 50, and h = 64
SOLUTION: Use the Law of Cosines to find the missing angle
measure.
Find the measure of the remaining angle.
Therefore, F
38 , G
51 and H
91 .
33. BCD, if B = 16 , c = 27, and d = 3
SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
eSolutions Manual - Powered by Cognero
Page 17
Find the measure of the remaining angle.
Find the measure of the remaining angle.
4-7 The Law of Sines and the Law of Cosines
Therefore, F
38 , G
51 and H
91 .
33. BCD, if B = 16 , c = 27, and d = 3
Therefore, C
162 , D
2 and b
24.1.
34. LMN, if l = 12, m = 4, and n = 9
SOLUTION: SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Cosines to find the missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Sines to find a missing angle
measure.
Find the measure of the remaining angle.
Therefore, C
162 , D
2 and b
24.1.
Find the measure of the remaining angle.
34. LMN, if l = 12, m = 4, and n = 9
SOLUTION: Use the Law of Cosines to find the missing angle
measure.
Therefore, L
131 , M
15 and N
34 .
35. AIRPLANES During her shift, a pilot flies from the
Columbus to Atlanta, a distance of 448 miles, and
then on to the Phoenix, a distance of 1583 miles.
From Phoenix, she returns home to Columbus, a
distance of 1667 miles. Determine the angles of the
triangle created by her flight path.
SOLUTION: Draw a diagram to represent the situation.
Use the Law of Sines to find a missing angle
measure.
Use the Law of Cosines to find an angle measure.
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Page 18
Use the Law of Sines to find a second angle
Find A.
Find the measure of the remaining angle.
4-7 The Law of Sines and the Law of Cosines
Therefore, L
131 , M
15 and N
34 .
35. AIRPLANES During her shift, a pilot flies from the
Therefore, the angles of the triangle created by the
flight path are about 15.6 , 71.5 , and 92.9 .
36. CATCH Lola rolls a ball on the ground at an angle Columbus to Atlanta, a distance of 448 miles, and
then on to the Phoenix, a distance of 1583 miles.
From Phoenix, she returns home to Columbus, a
distance of 1667 miles. Determine the angles of the
triangle created by her flight path.
of 23° to the right of her dog Buttons. If the ball rolls
a total distance of 48 feet, and she is standing 30 feet
away, how far will Buttons have to run to retrieve
the ball?
SOLUTION: Draw a diagram to model the situation.
Draw a diagram to represent the situation.
SOLUTION: Use the Law of Cosines to find x.
Use the Law of Cosines to find an angle measure.
Therefore, Buttons will have to run 23.5 feet to
retrieve that ball.
Use the Law of Sines to find a second angle
measure.
Use Heron’s Formula to find the area of each
triangle. Round to the nearest tenth.
37. x = 9 cm, y = 11 cm, z = 16 cm
SOLUTION: First, find the value of s.
Next, use Heron's Formula find the area of
.
Find A.
Therefore, the angles of the triangle created by the
flight path are about 15.6 , 71.5 , and 92.9 .
36. CATCH Lola rolls a ball on the ground at an angle of 23° to the right of her dog Buttons. If the ball rolls
a total distance of 48 feet, and she is standing 30 feet
away, how far will Buttons have to run to retrieve
the ball?
Therefore, the area of
2
is about 47.6 cm .
38. x = 29 in., y = 25 in., z = 27 in.
SOLUTION: First, find the value of s.
SOLUTION: Draw a diagram to model the situation.
Next, use Heron's Formula find the area of
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.
Page 19
4-7 Therefore,
The Lawtheofarea
Sines
of2Cosines
of andisthe
aboutLaw
47.6 cm
.
38. x = 29 in., y = 25 in., z = 27 in.
Therefore, the area of
41. x = 8 yd, y = 15 yd, z = 8 yd
SOLUTION: SOLUTION: First, find the value of s.
First, find the value of s.
Next, use Heron's Formula find the area of
Therefore, the area of
.
Next, use Heron's Formula find the area of
Therefore, the area of
SOLUTION: First, find the value of s.
First, find the value of s.
Next, use Heron's Formula find the area of
.
Next, use Heron's Formula find the area of
Therefore, the area of
2
is about 1133.0 ft .
40. x = 37 mm, y = 10 mm, z = 34mm
.
2
is about 2895.1 ft .
43. LANDSCAPING The Steele family want to expand their backyard by purchasing a vacant lot
adjacent to their property. To get a rough
measurement of the area of the lot, Mr. Steele
counted the steps needed to walk around the border
and diagonal of the lot.
SOLUTION: First, find the value of s.
Next, use Heron's Formula find the area of
41. x = 8 yd, y = 15 yd, z = 8 yd
2
is about 20.9 ft .
42. x = 133 mi, y = 82 mi, z = 77 mi
SOLUTION: Therefore, the area of
.
2
is about 312.2 cm .
39. x = 58 ft, y = 40 ft, z = 63 ft
Therefore, the area of
2
is about 167.6 ft .
.
2
is about 167.6 ft .
a. Estimate the entire area in steps.
b. Mr. Steele measured his step to be 1.8 feet.
Determine the area of the lot in square feet.
SOLUTION: SOLUTION: First, find the value of s.
a. Find the area of the Steele’s property. First, find s.
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Page 20
4-7 The Law of Sines and the Law of 2Cosines
Therefore, the area of
is about 2895.1 ft .
43. LANDSCAPING The Steele family want to expand their backyard by purchasing a vacant lot
adjacent to their property. To get a rough
measurement of the area of the lot, Mr. Steele
counted the steps needed to walk around the border
and diagonal of the lot.
Therefore, the area is about 14,617 square feet.
44. DANCE During a performance, a dancer remained
within a triangular area of the stage.
a. Find the area of stage used in the performance.
b. If the stage is 250 square feet, determine the
percentage of the stage used in the performance.
SOLUTION: a. Use Heron’s Formula to find the area. First find s.
a. Estimate the entire area in steps.
b. Mr. Steele measured his step to be 1.8 feet.
Determine the area of the lot in square feet.
SOLUTION: a. Find the area of the Steele’s property. First, find s.
Therefore, the area of stage used in the performance
is 163.9 square feet.
Use Heron's Formula find the area of the triangle.
Next, find the area of the vacant lot.
b. The stage is 250 square feet. So, 163.9 ÷ 250 =
0.656 or about 66% of the stage is used in the
performance
Find the area of each triangle to the nearest
tenth.
45. ABC, if A = 98 , b = 13 mm, and c = 8 mm
SOLUTION: Use Heron's Formula find the area of the triangle.
2
Therefore, the area of ABC is about 51.5 mm .
Therefore, the total area is 963.1 + 3548.4 or about
4511.5 square steps.
b. Use dimensional analysis to convert the area from
square steps to square feet.
46. JKL, if L = 67 , j = 11 yd, and k = 24 yd
SOLUTION: 2
Therefore, the area of ABC is about 121.5 mm .
Therefore, the area is about 14,617 square feet.
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44. DANCE
within a triangular area of the stage.
47. RST, if R = 35 , s = 42 ft, and t = 26 ft
SOLUTION: Page 21
of at least 75 square feet, what is the measure of the
third side?
4-7 The Law of Sines and the Law of Cosines
2
Therefore, the area of ABC is about 121.5 mm .
47. RST, if R = 35 , s = 42 ft, and t = 26 ft
SOLUTION: 2
Therefore, the area of ABC is about 313.2 mm .
48. XYZ, if Y = 124 , x = 16 m, and z = 18 m
SOLUTION: Draw a diagram to model the situation of a triangle
with side lengths of 15 and 18 feet. Because the 42º
angle is a nonincluded angle, there are two possible
triangles that can be formed.
Triangle 1
Use the Law of Sines to find C.
SOLUTION: 2
Therefore, the area of ABC is about 119.4 mm .
So, C = 180º – (42º + 53.41º) or about 84.59º. Use
the Law of Sines again to find c.
49. FGH, if F = 41 , g = 22 in., and h = 36 in.
SOLUTION: 2
Therefore, the area of ABC is about 259.8 mm .
Find the area of the triangle to make sure that it is at
least 75 square feet.
50. PQR, if Q = 153 , p = 27 cm, and r = 21 cm
SOLUTION: Triangle 2
Use the Law of Sines to find C.
2
Therefore, the area of ABC is about 128.7 mm .
51. DESIGN A free-standing art project requires a
triangular support piece for stability. Two sides of
the triangle must measure 18 and 15 feet in length
and a nonincluded angle must measure 42 . If
support purposes require the triangle to have an area
of at least 75 square feet, what is the measure of the
third side?
SOLUTION: Draw a diagram to model the situation of a triangle
with side lengths of 15 and 18 feet. Because the 42º
angle
is a nonincluded
angle, there are two possible
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triangles that can be formed.
So, C = 180º – (42º + 33.89º) or about 104.11º. Use
the Law of Sines again to find c.
Page 22
4-7 The Law of Sines and the Law of Cosines
Therefore, the measure of the third side is about 22.3
ft or about 26.1 ft.
So, C = 180º – (42º + 33.89º) or about 104.11º. Use
the Law of Sines again to find c.
Use Heron’s Formula to find the area of each
figure. Round answers to the nearest tenth.
Find the area of the triangle to make sure that it is at
least 75 square feet.
52. SOLUTION: Find the area of PQR. Because three side lengths
are given, you can use Heron’s formula. First, find s.
Therefore, the measure of the third side is about 22.3
ft or about 26.1 ft.
Use Heron’s Formula to find the area of each
figure. Round answers to the nearest tenth.
Find the area of
52. RSU. RU = 38.1 + 24.3 = 62.4.
SOLUTION: Find the area of
PQR. Because three side lengths
Find the area of
STV. SV = 34 + 43.2 = 77.2
Therefore, the total area is 808.2 + 1009.6 + 1143.2
= 2961.0 square meters.
53. SOLUTION: eSolutions Manual - Powered by Cognero
Page 23
First, find the area of EFJ. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
Therefore, the total area is 110.7 + 178 or about
288.7 square millimeters.
4-7
Therefore, the total area is 808.2 + 1009.6 + 1143.2
The
Law
of meters.
Sines and the Law of Cosines
=
2961.0
square
54. SOLUTION: First, find the area of ABD. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
53. SOLUTION: First, find the area of EFJ. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
Find the area of
Find the area of
BCD.
FGH.
Therefore, the total area is 15.0 + 28.3 or about 43.3
square centimeters.
Therefore, the total area is 110.7 + 178 or about
288.7 square millimeters.
55. SOLUTION: 54. SOLUTION: First, find the area of ABD. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
First, find the area of KLN. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
Find the area of
LMN.
Find the area of
BCD.
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Therefore, the total area is 15.0 + 28.3 or about 43.3
square centimeters.
Therefore, the total area 600 + 462.6 or about 1062.6
Page 24
square feet.
56. ZIP LINES A tourist attraction currently has its area isand
15.0the
+ 28.3
or about
43.3
4-7 Therefore,
The Lawtheoftotal
Sines
Law
of Cosines
square centimeters.
SOLUTION: Draw a diagram to represent the situation.
55. SOLUTION: First, find the area of KLN. Because three side
lengths are given, you can use Heron’s formula.
First, find s.
Find the area of
Recall from Geometry that when two parallel lines
are cut by a transversal, then consecutive angles are
supplementary.
LMN.
∠A + 39 + 72 = 180
∠A = 69
∠C = 72° − 31° = 41°
Therefore, the total area 600 + 462.6 or about 1062.6
square feet.
∠B = 180° − (69° + 41°) = 70°.
Use the Law of Sines to find a.
56. ZIP LINES A tourist attraction currently has its base connected to a tree platform 150 meters away
by a zip line. The owners now want to connect the
base to a second platform located across a canyon
and then connect the platforms to each other. The
bearings from the base to each platform and from
platform 1 to platform 2 are given. Find the distances
from the base to platform 2 and from platform 1 to
platform 2.
Use the Law of Sines again to find c.
SOLUTION: Draw a diagram to represent the situation.
Therefore, the distance from the base to platform 2
is about 149.02 meters, and the distance from
platform 1 to platform 2 is about 104.72 meters.
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Page 25
57. LIGHTHOUSES The bearing from the South Bay lighthouse to the Steep Rock lighthouse 25 miles
away is N 28 E. A small boat in distress spotted
4-7
Therefore, the distance from the base to platform 2
is
about
149.02
the the
distance
from
The
Law
ofmeters,
Sinesand
and
Law
of Cosines
platform 1 to platform 2 is about 104.72 meters.
Use the Law of Sines again to find the distance from
the lighthouse at Steep Rock to the boat.
57. LIGHTHOUSES The bearing from the South Bay lighthouse to the Steep Rock lighthouse 25 miles
away is N 28 E. A small boat in distress spotted
off the coast by each lighthouse has a bearing of N
50 W from South Bay and S 80 W from Steep
Rock. How far is each tower from the boat?
Therefore, South Bay is about 25.72 miles from the
boat and Steep Rock is about 31.92 miles from the
boat.
Find the area of each figure. Round answers to
the nearest tenth.
SOLUTION: Draw a diagram to represent the situation.
58. SOLUTION: First, find the area of the triangle with side lengths of
43 and 32 cm.
Recall from Geometry that when two parallel lines
are cut by a transversal, then alternate interior angles
are congruent. Therefore, C in ABC is 80 –
28 = 52 .
Next, find the area of the triangle with side lengths
of 43 and 51 cm.
B = 50 + 28 = 78
Therefore, the total area of the figure is 394.6 +
A = 180
– (52 + 78 ) = 50
2
548.3 or about 942.9 cm .
Use the Law of Sines to find the distance from the
lighthouse at South Bay to the boat.
59. SOLUTION: Use the Law of Sines again to find the distance from
the lighthouse at Steep Rock to the boat.
Each of the triangles that make up this composite
figure have one side length of 36 mm, a second side
length of 42 mm, and an included angle of 26°. Use
these values to find the area for one triangle.
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Page 26
Therefore, the total area of the figure is 394.6 +
Therefore, the total area of the figure is 2404.91 +
2
4-7 548.3
The or
Law
Sines
the Law of Cosines
aboutof942.9
cm and
.
2
1544.45 or about 3949.4 ft .
59. SOLUTION: Each of the triangles that make up this composite
figure have one side length of 36 mm, a second side
length of 42 mm, and an included angle of 26°. Use
these values to find the area for one triangle.
Therefore, the total area is 3(331.4) or about 994.2
2
mm .
61. SOLUTION: First, find the area of the triangle with side lengths of
32 and 37.5 in.
Next, find the area of the triangle with side lengths
of 32 and 42 in.
Therefore, the total area of the figure is 369.4 +
2
514.8 or about 884.2 in .
62. BRIDGE DESIGN In the figure below, 60. SOLUTION: First, find the area of the triangle with side lengths of
65 and 81 ft.
=
45 ,
= 55°, , B is the
midpoint of AC, and DE EG. If AD = 4 feet, DE
= 12 feet, and CE = 14 feet, find BF.
Next, find the area of the triangle with side lengths
of 67 and 87 ft.
SOLUTION: Therefore, the total area of the figure is 2404.91 +
Because AD = 4 and DE = 12, AE = 16. It is given
that CE = 14 and CED = 55 .
2
1544.45 or about 3949.4 ft .
Use the Law of Cosines to find AC.
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eSolutions
SOLUTION: First, find the area of the triangle with side lengths of
Page 27
Because B is the midpoint of AC, AB
6.98.
63. BUILDINGS Barbara wants to know the distance 4-7 The Law of Sines and the Law of Cosines
Use the Law of Cosines to find AC.
between the tops of two buildings R and S. On the
top of her building, she measures the distance
between the points T and U and finds the given angle
measures. Find the distance between the two
buildings.
Because B is the midpoint of AC, AB
6.98.
SOLUTION: Because two side lengths of ABC are now known,
the Pythagorean Theorem can be used to find BD.
Label the point at which
and
intersect as X.
Notice that FDG is a 45 -45 -90 triangle.
Because DE = 12 and DE ≅ EG, EG = 12. So, DG
= 24.
From the properties of 45 -45 -90 triangles, 24 =
So, x =
TXU is 180 − (65 + 38 ) = 77 . Use the Law
of Sines to find
Because BDA = 90 and FDE = 45 , BDF
= 45 . The measures of BD, DF, and BDF are
now known.
SXU = 180
XSU = 180
− 77 = 103 − (54 + 103 ) = 23
Use the Law of Sines to find
Use the Law of Cosines to find BF.
63. BUILDINGS Barbara wants to know the distance between the tops of two buildings R and S. On the
top of her building, she measures the distance
between the points T and U and finds the given angle
measures. Find the distance between the two
buildings.
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Because RXT and SXU are vertical angles,
RXT is 103 . Therefore, XRT is 180 − (53
+ 103 ) = 24 . Use the Law of Sines to find Page 28
4-7 Because
The LawRXT
of Sines
and the Law of Cosines
and SXU are vertical angles,
RXT is 103 . Therefore, XRT is 180 − (53
+ 103 ) = 24 . Use the Law of Sines to find
Therefore, the distance between the two buildings is
about 40.9 meters.
64. DRIVING After a high school football game, Della left the parking lot traveling 35 miles per hour in the
direction N 55 E. If Devon left 20 minutes after
Della at 45 miles per hour in the direction S 10 W,
how far apart are Devon and Della an hour and a
half after Della left?
Because ∠TXU and ∠RXS are vertical angles,
∠RXS is 77°. Use the Law of Cosines to find SOLUTION: Use the distance formula d = rt, where r is the rate
and t is time.
Therefore, the distance between the two buildings is
about 40.9 meters.
For Della, the distance traveled is d = 35(1.5) or 52.5
miles. Because Devon left 20 minutes after Della,
the time that he has been driving is
64. DRIVING After a high school football game, Della left the parking lot traveling 35 miles per hour in the
direction N 55 E. If Devon left 20 minutes after
Della at 45 miles per hour in the direction S 10 W,
how far apart are Devon and Della an hour and a
half after Della left?
or about 1.17 hours. So, for Devon, the distance traveled is d ≈ 45(1.17) or
about 52.5 miles.
Because Della is traveling N 55º E and Devon is
traveling S 10º W, the angle between them is 35° +
90° + 10° or 135°, as shown.
SOLUTION: Use the distance formula d = rt, where r is the rate
and t is time.
For Della, the distance traveled is d = 35(1.5) or 52.5
miles. Because Devon left 20 minutes after Della,
the time that he has been driving is
Use the Law of Cosines to find the distance
between them.
or about 1.17 hours. So, for Devon, the distance traveled is d ≈ 45(1.17) or
about 52.5 miles.
Because Della is traveling N 55º E and Devon is
traveling S 10º W, the angle between them is 35° +
90° + 10° or 135°, as shown.
Therefore, after an hour and a half, Devon and Della
are about 97 miles apart.
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65. ERROR ANALYSIS Monique and Rogelio are
Page 29
solving an acute triangle in which
= 34 , a =
16, and b = 21. Monique thinks that the triangle has
hour and
Della
4-7 Therefore,
The Lawafter
of an
Sines
anda half,
the Devon
Law and
of Cosines
are about 97 miles apart.
65. ERROR ANALYSIS Monique and Rogelio are
solving an acute triangle in which
= 34 , a =
16, and b = 21. Monique thinks that the triangle has
one solution, while Rogelio thinks that the triangle
has no solution. Is either of them correct? Explain
your reasoning.
The Pythagorean Theorem should be used to find a
missing side length of a right triangle when given two
side lengths. However, the Pythagorean Theorem
can only be used to solve for the missing side. For
example, use the Pythagorean Theorem to find x in
the in ΔABC.
SOLUTION: This problem involves the ambiguous case. There
may be 0, 1, or 2 solutions. Compare a with b and h.
For the acute case, h = 21 sin 34 or 11.7. Because
a < b and h < a, there are two solutions. Therefore,
neither Monique nor Rogelio are correct.
The trigonometric ratios can be used to solve a right
triangle when given two side lengths or one side
length and the measure of an acute angle of the
triangle. For example, you can use trigonometry to
find the measure of ∠A in ΔABC.
66. Writing in Math Explain the different
circumstances in which you would use the Law of
Cosines, the Law of Sines, the Pythagorean
Theorem, and the trigonometric ratios to solve a
triangle.
SOLUTION: The Law of Cosines should be used to solve an
oblique triangle when given three side lengths or two
side lengths and the included angle. For example, use
the Law of Cosines to find x in ΔABC. The Law of Sines should be used to solve an oblique
triangle when given the measures of two angles and
a noninlcuded side, two angles and the included side,
or two sides and a nonincluded angle. For example,
use the Law of Sines to find x in ΔQRS.
The Pythagorean Theorem should be used to find a
missing side length of a right triangle when given two
side lengths. However, the Pythagorean Theorem
can only be used to solve for the missing side. For
example, use the Pythagorean Theorem to find x in
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the in
ΔABC.
67. REASONING Why does an obtuse measurement
appear on the graphing calculator for inverse cosine
while negative measures appear for inverse sine?
SOLUTION: The range of sine is [–1, 1], which is included in the
domain of [−90, 90] or [− , ]. Since the range
includes negative numbers, it is possible for the
inverse sign to be negative.
The range of cosine is [–1, 1], which is included in
the domain of [0, 180] or [0, ]. Since the domain is
from [0, 180] it is possible for there to be obtuse
measurements.
68. PROOF Show for a given rhombus with a side
Page 30
length of s and an included angle of θ that the area
2
can be found with the formula A = s sin θ.
4-7 The Law of Sines and the Law of Cosines
68. PROOF Show for a given rhombus with a side
for one of the triangles is
2
s sin
. To find the
area of the rhombus, double the area for one
2
triangle. So, the area of rhombus ABCD is s sin
.
69. PROOF Derive the Law of Sines.
length of s and an included angle of θ that the area
SOLUTION: 2
can be found with the formula A = s sin θ.
SOLUTION: Draw rhombus ABCD with side length of s.
Sample answer: Let h 1 be an altitude of either
triangle shown above. From the definition of the sine
function, h 1 = b sin A, and h 1 = a sin B. Therefore, b
sin A = a sin B.
A line drawn from B to D creates two congruent
triangles. Using the area formula for SAS, the area
for one of the triangles is
2
s sin
0 and sin B
. To find the
area of the rhombus, double the area for one
2
triangle. So, the area of rhombus ABCD is s sin
= Dividing by ab yields
, where sin A ≠
0. .
When an altitude h 2 is drawn from vertex B to side
AC (extended in the obtuse triangle), h 2 = c sin A,
69. PROOF Derive the Law of Sines.
and h 2 = a sin C. SOLUTION: Therefore, c sin A = a sin C.
= Dividing by ac yields
. Sample answer: Let h 1 be an altitude of either
triangle shown above. From the definition of the sine
function, h 1 = b sin A, and h 1 = a sin B. Therefore, b
sin A = a sin B.
Dividing by ab yields
0 and sin B
= , where sin A ≠
By the Transitive Property of equality,
= = .
70. PROOF Consider the figure below.
0. When an altitude h 2 is drawn from vertex B to side
AC (extended in the obtuse triangle), h 2 = c sin A,
and h 2 = a sin C. 2
Therefore, c sin A = a sin C.
= 2
2
2
2
formula a = b + c − 2bc cos A in the Law of
Cosines.
* Use the Pythagorean Theorem for DBC.
Dividing by ac yields
a. Use the figure and hints below to derive the first
2
* In ADB, c = x + h .
. * cos A =
By the Transitive Property of equality,
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= .
= b. Explain how you would go about deriving the
other two formulas in the Law of Cosines.
SOLUTION: a.
First use the Pythagorean Theorem on ΔDCB.
Page 31
= By the Transitive Property of equality,
4-7 The = Law of. Sines and the Law of Cosines
above Mars and is now positioned directly above one
of the poles. The radius of Mars is 2110 miles. If the
satellite was positioned at point X 14 minutes ago,
approximately how many hours does it take for the
satellite to complete a full orbit, assuming that it
travels at a constant rate around a circular orbit?
70. PROOF Consider the figure below.
a. Use the figure and hints below to derive the first
2
2
2
2
2
SOLUTION: formula a = b + c − 2bc cos A in the Law of
Cosines.
* Use the Pythagorean Theorem for DBC.
Draw a diagram to represent the situation. Let B
represent the location of the satellite when directly
above the pole, let A represent a point directly below
B on the surface of Mars, and let C represent the
center of Mars.
2
* In ADB, c = x + h .
* cos A =
b. Explain how you would go about deriving the
other two formulas in the Law of Cosines.
SOLUTION: a.
First use the Pythagorean Theorem on ΔDCB.
2
2
a 2 = (b − x) + h Pyth Thm.
Foil and simplify. 2
2
2
2
2
2
2
a 2 = b − 2bx + x + h Expand (b − x) + h .
2
2
2
Substitute x + h . for c .
2
2
2
a 2 = b − 2bx + c c = x + h .
Substitute c cos A for x. 2
2
a 2 = b − 2b(c cos A) + c x = c cos A.
Then simplify. 2
2
a 2 = b + c − 2bc cos A Comm. Prop.
b. To solve for b 2 and c2, altitudes can be drawn
from A and C and the same process can be used.
To find the amount of time that it takes for the
satellite to complete one full orbit, start by finding the
measure of the arc intercepted by points B and X.
Use the information given to find the measure of
The length of XC is 2110 + 850 or 2960 miles.
Two side lengths and an angle measure are now
known.
71. CHALLENGE A satellite is orbiting 850 miles
above Mars and is now positioned directly above one
of the poles. The radius of Mars is 2110 miles. If the
satellite was positioned at point X 14 minutes ago,
approximately how many hours does it take for the
satellite to complete a full orbit, assuming that it
travels at a constant rate around a circular orbit?
Use the Law of Sines to find
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Page 32
4-7 The Law of Sines and the Law of Cosines
Therefore, one full orbit takes approximately 4.36
hours or 4 hours and 22 minutes.
Use the Law of Sines to find
72. Writing in Math Describe why solving a triangle in
which h < a < b using the Law of Sines results in
two solutions. Is this also true when using the Law
of Cosines? Explain your reasoning.
SOLUTION: On the unit circle, the sine function is positive in the
first two quadrants, or when 0 < θ < π. Additionally,
if sin = x, there also exists sin (180 − ) = x. This
suggests that there will be two possible values of θ
So,
is 180º – (125º + 35.73º) or about 19.27º.
when finding sin
Because point B is directly above point A,
From geometry, −1
x=
. =
. Therefore,
≈ 19.27 . Find the amount
of time that it takes to complete one full orbit or 360º.
Convert 261.55 minutes to hours.
This does not apply to the Law of Cosines. The cosine function is only positive on − < < .
Because the measure of an angle of a triangle must
be greater than zero, there is only one value for .
Therefore, one full orbit takes approximately 4.36
hours or 4 hours and 22 minutes.
Find the exact value of each expression, if it
exists.
72. Writing in Math Describe why solving a triangle in
which h < a < b using the Law of Sines results in
two solutions. Is this also true when using the Law
of Cosines? Explain your reasoning.
SOLUTION: SOLUTION: On the unit circle, the sine function is positive in the
first two quadrants, or when 0 < θ < π. Additionally,
if sin = x, there also exists sin (180 − ) = x. This
suggests that there will be two possible values of θ
when finding sin
73. −1
x=
Find a point on the unit circle on the interval
with a x-coordinate of
.
. This does not apply to the Law of Cosines. The cosine function is only positive on − < < .
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Because the measure of an angle of a triangle must
When t =
.
, cos t =
. Therefore, cos
–1
=
Page 33
This does not apply to the Law of Cosines. The cosine function is only positive on − < < .
When t =
the measure
of an
angle
of Law
a triangle
4-7 Because
The Law
of Sines
and
the
of must
Cosines
be greater than zero, there is only one value for
. Therefore, sin
–1
= .
.
Find the exact value of each expression, if it
exists.
, sin t =
75. arctan 1
SOLUTION: 73. Find a point on the unit circle on the interval
such that SOLUTION: =1.
Find a point on the unit circle on the interval
with a x-coordinate of
.
When t =
When t =
, cos t =
. Therefore, cos
–1
=
, tan t =
. Therefore, tan
–1
1=
.
.
76. sin−1
SOLUTION: 74. sin−1
Find a point on the unit circle on the interval
SOLUTION: with a y-coordinate of
.
Find a point on the unit circle on the interval
with a y-coordinate of
.
When t =
When t =
, sin t =
. Therefore, sin
–1
.
75. arctan 1
SOLUTION: eSolutions Manual - Powered by Cognero
Find a point on the unit circle on the interval
such that =1.
= , sin t =
. Therefore, sin
–1
= .
Identify the damping factor f(x )of each function.
Then use a graphing calculator to sketch the
graphs of f (x), −f (x), and the given function in
the same viewing window. Describe the
Page 34
behavior of the graph.
77. y = −2x sin x
SOLUTION: When t =
, sin t =
. Therefore, sin
–1
= 4-7 The
. Law of Sines and the Law of Cosines
Identify the damping factor f(x )of each function.
Then use a graphing calculator to sketch the
graphs of f (x), −f (x), and the given function in
the same viewing window. Describe the
behavior of the graph.
77. y = −2x sin x
The amplitude of the function is decreasing as x
approaches 0.
79. y = (x – 1)2 sin x
SOLUTION: 2
The function y = (x – 1) sin x is the product of the
2
functions y = (x – 1) and y = sin x, so f (x) = (x – 1)
2
.
SOLUTION: The function y = −2x sin x is the product of the
functions y = −2x and y = sin x, so f (x) = −2x.
The amplitude of the function is decreasing as x
approaches 0.
The amplitude of the function is decreasing as x
approaches 0.
80. y = −4x2 cos x
SOLUTION: 2
78. y =
The function y = −4x cos x is the product of the
2
2
functions y = −4x and y = cos x, so f (x) = −4x .
x cos x
SOLUTION: The function y =
functions y =
x cos x is the product of the
x and y = cos x, so f (x) =
x.
The amplitude of the function is decreasing as x
approaches 0.
81. CARTOGRAPHY The distance around Earth The amplitude of the function is decreasing as x
approaches 0.
79. y = (x – 1)2 sin x
SOLUTION: 2
The function y = (x – 1) sin x is the product of the
2
functions y = (x – 1) and y = sin x, so f (x) = (x – 1)
2
.
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The amplitude of the function is decreasing as x
approaches 0.
along a given latitude can be found using C =
, where r is the radius of Earth and L is the
latitude. The radius of Earth is approximately 3960
miles. Make a table of values for the latitude and
corresponding distance around Earth that includes L
= 0 , 30 , 45 , 60 , and 90 . Use the table to
describe the distances along the latitudes as you go
from 0 at the equator to 90 at a pole.
SOLUTION: Substitute 0 , 30 , 45 , 60 , and 90 for L in C =
2pr cos L, to determine the corresponding distance.
Latitude
Distance
0º
24,881.4
30º
21,547.9
45º
17,593.8
60º
12,440.7
90º
0
Page 35
The distances range from about 24,881 miles to 0
miles.
x
4-7
The
the function
decreasing
as xCosines
Theamplitude
Law ofofSines
andisthe
Law of
approaches 0.
The equation is approximately y = 1.0091(0.9805) .
b. Let eu = 0.9805. Solve for u.
81. CARTOGRAPHY The distance around Earth along a given latitude can be found using C =
, where r is the radius of Earth and L is the
latitude. The radius of Earth is approximately 3960
miles. Make a table of values for the latitude and
corresponding distance around Earth that includes L
= 0 , 30 , 45 , 60 , and 90 . Use the table to
describe the distances along the latitudes as you go
from 0 at the equator to 90 at a pole.
Substitute e
equation.
−0.0197
for 0.9805 in the regression
SOLUTION: Substitute 0 , 30 , 45 , 60 , and 90 for L in C =
2pr cos L, to determine the corresponding distance.
Latitude
Distance
0º
24,881.4
30º
21,547.9
45º
17,593.8
60º
12,440.7
90º
0
The distances range from about 24,881 miles to 0
miles.
c. From the regression equation
,
the value of N 0 is 1.0091 and the value of N should
be 0.01. Solve for x.
82. RADIOACTIVITY A scientist starts with a 1gram sample of lead-211. The amount of the sample
remaining after various times is shown in the table
below.
a. Find an exponential regression equation for the
amount y of lead as a function of time x.
b. Write the regression equation in terms of base e.
c. Use the equation from part b to estimate when
there will be 0.01 gram of lead-211 present.
SOLUTION: a. Enter the data into the calculator and then select
ExpReg.
There will be 0.01 gram remaining after about 234
minutes.
Write a polynomial function of least degree with
real coefficients in standard form that has the
given zeros.
83. –1, 1, 5
SOLUTION: Using the Linear Factorization Theorem and the
zeros −1, 1, and 5, write f (x) as follows.
f(x) = a[x − (–1)][x − (1)][x − (5)]
Let a = 1. Then write the function in standard form.
x
The equation is approximately y = 1.0091(0.9805) .
b. Let eu = 0.9805. Solve for u.
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Therefore, a function of least degree that has −1, 1,
3
2
and 5, as zeros is f (x) = x – 5x – x + 5 or any nonzero multiple of f (x).
Page 36
84. –2, −0.5, 4
will be of
0.01Sines
gram remaining
about
4-7 There
The Law
and theafter
Law
of 234
Cosines
minutes.
Write a polynomial function of least degree with
real coefficients in standard form that has the
given zeros.
83. –1, 1, 5
SOLUTION: Using the Linear Factorization Theorem and the
zeros −1, 1, and 5, write f (x) as follows.
f(x) = a[x − (–1)][x − (1)][x − (5)]
Let a = 1. Then write the function in standard form.
Therefore, a function of least degree that has −1, 1,
3
2
and 5, as zeros is f (x) = x – 5x – x + 5 or any nonzero multiple of f (x).
Multiply the polynomial by 2 so that the coefficient of
2
the x -term is an integer. Therefore, a function of
least degree that has −2, –0.5, and 4, as zeros is f (x)
3
2
= 2x – 3x – 18x – 8or any nonzero multiple of f (x).
85. −3, −2i, 2i
SOLUTION: Using the Linear Factorization Theorem and the
zeros –3, 2i, and −2i, write f (x) as follows.
f(x) = a[x − (–3)] [x − (2i)][x − (−2i)]
Let a = 1. Then write the function in standard form.
Therefore, a function of least degree that has –3, 2i,
3
2
and −2i as zeros is f (x) = x + 3x + 4x + 12or any
nonzero multiple of f (x).
86. –5i, −i, i, 5i
SOLUTION: 84. –2, −0.5, 4
SOLUTION: Using the Linear Factorization Theorem and the
zeros −2, –0.5, and 4, write f (x) as follows.
f(x) = a[x − (–2)][x − (–0.5)][x − (4)]
Let a = 1. Then write the function in standard form.
Using the Linear Factorization Theorem and the
zeros –5i , – i , i, and 5i, write f (x) as follows.
f(x) = a [x − (–5i)][x − (−i)][x − (i)][x − (5i)]
Let a = 1. Then write the function in standard form.
Therefore, a function of least degree that has –5i , –
4
2
i , i, and 5i as zeros is f (x) = x + 26x + 25or any
nonzero multiple of f (x).
Multiply the polynomial by 2 so that the coefficient of
2
the x -term is an integer. Therefore, a function of
least degree that has −2, –0.5, and 4, as zeros is f (x)
3
87. SAT/ACT Which of the following is the perimeter
of the triangle shown?
2
= 2x – 3x – 18x – 8or any nonzero multiple of f (x).
85. −3, −2i, 2i
SOLUTION: Using the Linear Factorization Theorem and the
zeros –3, 2i, and −2i, write f (x) as follows.
f(x) = a[x − (–3)] [x − (2i)][x − (−2i)]
Let a = 1. Then write the function in standard form.
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A 49.0 cm
B 66.0 cm
C 71.2 cm
D 91.4 cm
E 93.2 cm
SOLUTION: Page 37
When the altitude is drawn on the triangle, it is
separated into two congruent right triangles.
Because an acute angle and the opposite side length
4-7
Therefore, a function of least degree that has –5i , –
4
2
iThe
, i, and
5i asof
zeros
is f (x)and
= x the
+ 26xLaw
+ 25or
Law
Sines
ofany
Cosines
nonzero multiple of f (x).
87. SAT/ACT Which of the following is the perimeter
of the triangle shown?
A 49.0 cm
B 66.0 cm
C 71.2 cm
D 91.4 cm
E 93.2 cm
The perimeter of the triangle is 22 + 2(35.6) ≈ 93.2 cm. Therefore, the correct answer is E.
88. In
DEF, what is the value of θ to the nearest
degree?
F 26
G 74
H 80
J 141
SOLUTION: Use the Law of Cosines to find θ.
SOLUTION: When the altitude is drawn on the triangle, it is
separated into two congruent right triangles.
Because an acute angle and the opposite side length
are given, you can use the sine function to find the
length of the hypotenuse.
Therefore, the correct answer is G.
89. FREE RESPONSE The pendulum below moves
according to θ =
The perimeter of the triangle is 22 + 2(35.6) ≈ 93.2 cm. Therefore, the correct answer is E.
88. In
DEF, what is the value of θ to the nearest
degree?
cos 12t, where θ is the angular
displacement in radians and t is the time in seconds.
a. Set the mode to radians and graph the function
for 0 ≤ t ≤ 2.
b. What are the period, amplitude, and frequency of
the function? What do they mean in the context of
this situation?
c. What is the maximum angular displacement of the
pendulum in degrees?
d. What does the midline of the graph represent?
e . At what times is the pendulum displaced 5
degrees?
SOLUTION: F 26
G 74
H 80
J 141
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SOLUTION: Use the Law of Cosines to find θ.
a.
Page 38
4-7
pendulum in degrees?
d. What does the midline of the graph represent?
e . At what times is the pendulum displaced 5
degrees?
The Law of Sines and the Law of Cosines
SOLUTION: a.
e . Draw a diagram of the situation.
d. The midline represents when the pendulum is
vertical and there is no angular displacement.
b. Period:
Convert 5º to radians.
Determine the time θ =
on 0 < t < 0.524.
The period represents the amount of time that it
takes for the pendulum to complete one full swing or
cycle.
Amplitude:
The amplitude represents the maximum angular
displacement of the pendulum.
Frequency:
So, the pendulum is displaced 5º after 0.101 second.
Determine the time when the pendulum is vertical.
The frequency represents the number of swings or
cycles that the pendulum completes per second.
c. The maximum angular displacement occurs when
θ=
The pendulum is vertical when t =
d. The midline represents when the pendulum is
vertical and there is no angular displacement.
e . Draw a diagram of the situation.
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or about
0.1309 second, which is 0.0299 second after the
pendulum was displaced 5 the first time. So, the
pendulum is displaced 5 for a second time at
0.1608 second.
Therefore, in general, the pendulum is displaced 5º at
about 0.101 + 0.524n seconds and 0.161 + 0.524n
Page 39
seconds.
0.1309 second, which is 0.0299 second after the
pendulum was displaced 5 the first time. So, the
pendulum is displaced 5 for a second time at
0.1608 second.
4-7 The Law of Sines and the Law of Cosines
Therefore, in general, the pendulum is displaced 5º at
about 0.101 + 0.524n seconds and 0.161 + 0.524n
seconds.
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Page 40