4-7 The Law of Sines and the Law of Cosines
Solve each triangle. Round to the nearest tenth, if necessary.
1.
ANSWER:
A = 32 , a = 11.2, b = 19.8
2.
ANSWER:
H = 15 , f = 55.5, g = 64.5
3.
ANSWER:
K = 82 , j = 16.2, ℓ = 21.4
4.
ANSWER:
S = 62 , r = 6.6, s = 7
5.
ANSWER:
T = 20 , t = 29.0, u = 17.7
6.
ANSWER:
D = 83 , b = 12.8, c = 28.7
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Page 1
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.
ANSWER:
4-7 The Law of Sines and the Law of Cosines
T = 20 , t = 29.0, u = 17.7
6.
ANSWER:
D = 83 , b = 12.8, 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?
ANSWER:
about 0.85 ft
8. ARCHITECTURE An architect’s client wants to build a home based on the architect Jon Lautner’s SheatsGoldstein 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.
ANSWER:
left: about 20.1 ft, right: about 49.2 ft; 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
ANSWER:
a. about 138.4 mi
b. about 135.9 mi
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
ANSWER:
B = 48 , C = 24 , c = 3.9
11. a = 14, b = 15, A = 117
ANSWER:
no solution
12. a = 18, b = 12, A = 27
ANSWER:
B = Manual
18 , C- Powered
= 135 by
, cCognero
= 27.8
eSolutions
13. a = 35, b = 24, A = 92
Page 2
11. a = 14, b = 15, A = 117
4-7 ANSWER:
The Law of Sines and the Law of Cosines
no solution
12. a = 18, b = 12, A = 27
ANSWER:
B = 18 , C = 135 , c = 27.8
13. a = 35, b = 24, A = 92
ANSWER:
B = 43 , C = 45 , c = 24.7
14. a = 14, b = 6, A = 145
ANSWER:
B = 14 , C = 21 , c = 8.7
15. a = 19, b = 38, A = 30
ANSWER:
B = 90 , C = 60 , c = 32.9
16. a = 5, b = 6, A = 63
ANSWER:
no solution
17. a = 10, b =
, A = 45
ANSWER:
B = 90 , C = 45 , c = 10
18. SKIING A ski lift rises at a 28 angle during the first 20 feet up a mountain to achieve a height of 41 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.
ANSWER:
about 41 ft
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
ANSWER:
B = 63 , C = 78 , c = 18.7 and B = 117 , C = 24 , c = 7.8
20. A = 26 , a = 5, b = 9
ANSWER:
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B = 52 , C = 102 , c = 11.2 and B = 128 , C = 26 , c = 5.0
21. A = 61 , a = 14, b = 15
Page 3
19. A = 39 , a = 12, b = 17
ANSWER:
4-7 BThe
and
= 63Law
, C =of
78 Sines
, c = 18.7
andthe
B = Law
117 , of
C =Cosines
24 , c = 7.8
20. A = 26 , a = 5, b = 9
ANSWER:
B = 52 , C = 102 , c = 11.2 and B = 128 , C = 26 , c = 5.0
21. A = 61 , a = 14, b = 15
ANSWER:
B = 70 , C = 49 , c = 12.2 and B = 110 , C = 9 , c = 2.5
22. A = 47 , a = 25, b = 34
ANSWER:
B = 84 , C = 49 , c = 25.8 and B = 96 , C = 37 , c = 20.6
23. A = 54 , a = 31, b = 36
ANSWER:
B = 70°, C = 56°, c = 31.8 and B = 110°, C = 16°, c = 10.6
24. A = 18 , a = 8, b = 13
ANSWER:
B = 30 , C = 132 , c = 19.2 and B = 150 , C = 12 , c = 5.4
25. BROADCASTING A radio tower located 38 miles along Industrial Parkway transmits radio broadcasts over a 30mile radius. Industrial Parkway intersects the interstate at a 41º angle. How far along the interstate can vehicles pick
up the broadcasting signal?
ANSWER:
33.4 mi
26. BOATING The light from a lighthouse can be seen from an 18-mile radius. A boat is positioned 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.
ANSWER:
18.3 mi
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
ANSWER:
B = 39 , C = 99 , a = 12.9
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28. XYZ, if x = 5, y = 18, and z = 14
Page 4
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.
4-7 ANSWER:
The Law of Sines and the Law of Cosines
18.3 mi
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
ANSWER:
B = 39 , C = 99 , a = 12.9
28. XYZ, if x = 5, y = 18, and z = 14
ANSWER:
X = 11 , Y = 137 , Z = 32
29.
PQR, if P = 73 , q = 7, and r = 15
ANSWER:
Q = 27 , R = 80 , p = 14.6
30.
JKL, if J = 125 , k = 24, and l = 33
ANSWER:
K = 23 , L = 32 , j = 50.7
31.
RST, if r = 35, s = 22, and t = 25
ANSWER:
R = 96 , S = 39 , T = 45
32.
FGH, if f = 39, g = 50, and h = 64
ANSWER:
F = 38 , G = 52 , H = 90
33.
BCD, if B = 16 , c = 27, and d = 3
ANSWER:
C = 162 º, D = 2 , b = 24.1
34.
LMN, if
= 12, m = 4, and n = 9
ANSWER:
L = 131 , M = 15 , 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.
ANSWER:
15.6 ; 71.5 ; 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?
ANSWER:
23.5 ft
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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
Page 5
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.
ANSWER:
4-7 The Law of Sines and the Law of Cosines
15.6 ; 71.5 ; 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?
ANSWER:
23.5 ft
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
ANSWER:
47.6 cm
2
38. x = 29 in., y = 25 in., z = 27 in.
ANSWER:
312.2 in
2
39. x = 58 ft, y = 40 ft, z = 63 ft
ANSWER:
1133.0 ft
2
40. x = 37 mm, y = 10 mm, z = 34mm
ANSWER:
167.6 mm
2
41. x = 8 yd, y = 15 yd, z = 8 yd
ANSWER:
2
20.9 yd
42. x = 133 mi, y = 82 mi, z = 77 mi
ANSWER:
2895.1 mi
2
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.
a. Estimate the area of the lot in steps.
b. Mr. Steele measured his step to be 1.8 feet. Determine the area of the lot in square feet.
ANSWER:
eSolutions
Manual - Powered by Cognero
a. 4511.5 steps
b. 14,617 ft2
2
Page 6
42. x = 133 mi, y = 82 mi, z = 77 mi
ANSWER:
4-7 The Law
2 of Sines and the Law of Cosines
2895.1 mi
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.
a. Estimate the area of the lot in steps.
b. Mr. Steele measured his step to be 1.8 feet. Determine the area of the lot in square feet.
ANSWER:
a. 4511.5 steps
b. 14,617 ft2
2
44. DANCE During a performance, a dancer remained within a triangular area of the stage.
a. Determine 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.
ANSWER:
a. 163.9 ft
b. 66%
2
Find the area of each triangle to the nearest tenth.
45. ABC, if A = 98 , b = 13 mm, and c = 8 mm
ANSWER:
2
51.5 mm
46.
JKL, if L = 67 , j = 11 yd, and k = 24 yd
ANSWER:
2
121.5 yd
47.
RST, if R = 35 , s = 42 ft, and t = 26 ft
ANSWER:
313.2 ft
2
48. XYZ, if Y = 124 , x = 16 m, and z = 18 m
ANSWER:
119.4 m
2
49. FGH,
if -FPowered
= 41 ,bygCognero
= 22 in.,
eSolutions
Manual
ANSWER:
2
and h = 36 in.
Page 7
48. XYZ, if Y = 124 , x = 16 m, and z = 18 m
ANSWER:
4-7 The Law
of Sines and the Law of Cosines
2
119.4 m
49.
FGH, if F = 41 , g = 22 in., and h = 36 in.
ANSWER:
259.8 in
50.
2
PQR, if Q = 153 , p = 27 cm, and r = 21 cm
ANSWER:
128.7 cm
2
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 150 square feet, what is the measure of the third side?
ANSWER:
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.
52.
ANSWER:
2961.0 m
2
53.
ANSWER:
288.7 mm
2
54.
ANSWER:
43.3 cm
2
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Page 8
53.
ANSWER:
4-7 The Law
2 of Sines and the Law of Cosines
288.7 mm
54.
ANSWER:
43.3 cm
2
55.
ANSWER:
1062.6 ft
2
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.
ANSWER:
about 149.02 m, about 104.72 m
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?
ANSWER:
South Bay is about 25.72 mi from the boat and Steep Rock is about 31.92 mi from the boat.
Find the area of each figure. Round answers to the nearest tenth.
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ANSWER:
4-7 The Law of Sines and the Law of Cosines
South Bay is about 25.72 mi from the boat and Steep Rock is about 31.92 mi from the boat.
Find the area of each figure. Round answers to the nearest tenth.
58.
ANSWER:
942.9 cm
2
59.
ANSWER:
994.2 mm
2
60.
ANSWER:
2
3949.4 ft
61.
ANSWER:
884.2 in
2
62. BRIDGE DESIGN the figure below,
DE
= 45 ,
= 55°,
EG. If AD = 4 feet, DE = 12 feet, and CE = 14 feet, find BF.
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ANSWER:
, B is the midpoint of AC, and
Page 10
61.
ANSWER:
4-7 The Law
of Sines and the Law of Cosines
2
884.2 in
62. BRIDGE DESIGN the figure below,
DE
= 45 ,
= 55°,
EG. If AD = 4 feet, DE = 12 feet, and CE = 14 feet, find BF.
, B is the midpoint of AC, and
ANSWER:
13.5 ft
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.
ANSWER:
40.9 m
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?
ANSWER:
97 mi
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.
ANSWER:
Neither; for the acute case, h = 21 sin 34 or 11.7. Because a < b and h < a, there are two solutions.
66. Writing in Math Explain the different circumstances in which you would use the Law of Cosines, the Law of
eSolutions
Manual
- Powered by Cognero
Sines,
the Pythagorean
Theorem,
ANSWER:
and the trigonometric ratios to solve a triangle.
Page 11
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.
ANSWER:
4-7 The Law of Sines and the Law of Cosines
Neither; for the acute case, h = 21 sin 34 or 11.7. Because a < b and h < a, there are two solutions.
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.
ANSWER:
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. 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. 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. 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.
67. REASONING Why does an obtuse measurement appear on the graphing calculator for inverse cosine while
negative measures appear for inverse sine?
ANSWER:
Sample answer: The domain of inverse cosine includes angles measured 0 to 180 degrees. The domain of inverse
sine includes angles measured −90 to 90 degrees.
68. PROOF Show for a given rhombus with a side length of s and an included angle of θ that the area can be found
2
with the formula A = s sin θ.
ANSWER:
Refer to rhombus ABCD.
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
(s)(s)sin θ or
2
s sin
. To find the area of the rhombus, double the area for one triangle. So, the
2
area of rhombus ABCD is
or s sin
.
69. PROOF Derive the Law of Sines using the area ofa triangle formula
ANSWER:
Sample answer: Let h 1 be an altitude of either triangle shown above. From the definition of the sine function, sin A =
or h 1 = b sin A, and sin B =
, where sin A ≠ 0 and sin B
eSolutions Manual - Powered by Cognero
obtuse triangle), sin A =
or h 1 = a sin B. Equating the two values of h 1, b sin A = a sin B, or
=
0. When an altitude h 2 is drawn from vertex B to side AC (extended in thePage 12
or h 2 = c sin A, and sin C =
or h 2 = a sin C. Equating the two values of h 2, c sin A
(s)(s)sin θ or
triangles is
2
s sin
. To find the area of the rhombus, double the area for one triangle. So, the
2
4-7 area
TheofLaw
of ABCD
Sinesisand the Law
Cosines
rhombus
or sofsin
.
69. PROOF Derive the Law of Sines using the area ofa triangle formula
ANSWER:
Sample answer: Let h 1 be an altitude of either triangle shown above. From the definition of the sine function, sin A =
or h 1 = b sin A, and sin B =
, where sin A ≠ 0 and sin B
=
=
0. When an altitude h 2 is drawn from vertex B to side AC (extended in the
or h 2 = c sin A, and sin C =
obtuse triangle), sin A =
= a sin C, or
or h 1 = a sin B. Equating the two values of h 1, b sin A = a sin B, or
or h 2 = a sin C. Equating the two values of h 2, c sin A
. By the Transitive Property of equality,
=
=
.
70. PROOF Consider the figure below.
a. Use the figure and hints below to derive the first formula a 2 = b 2 + c2 − 2bc cos A in the Law of Cosines.
* Use the Pythagorean Theorem for DBC.
2
2
2
* In ADB, c = x + h .
* cos A =
b. Explain how you would go about deriving the other two formulas, b 2 = a 2 + c2 − 2ac cos B and c2 = a 2 + b 2 −
2ab cos C, in the Law of Cosines.
ANSWER:
a.
2
2
a 2 = (b − x) + h
2
2
2
= b − 2bx + x + h
2
= b − 2bx + c
Use the Pythagorean Theorem for
2
2
Expand (b − x) + h .
2
2
2
2
= b + c − 2bc cos A
2
2
In ADB, c = x + h .
= b − 2b(c cos A) + c
2
2
DBC.
2
cos A =
, so x = c cos A.
Commutative Property
2
b. To solve for b and c , altitudes can be drawn from A and C and the same process can be used.
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
to complete a full orbit, assuming that it travels at a constant rate around a circular
Page 13
orbit?
hours
does- it
take for
the satellite
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by Cognero
2
= b − 2b(c cos A) + c
2
2
cos A =
2
= b + c − 2bc cos A
, so x = c cos A.
Commutative Property
4-7 The Law of 2Sines2and the Law of Cosines
b. To solve for b and c , altitudes can be drawn from A and C and the same process can be used.
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?
ANSWER:
about 4.36 h or 4 h 22 min
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.
ANSWER:
Sample answer: On the unit circle, the sine function is positive in the first two quadrants, or when 0 < < .
Additionally, if sin = x, there also exists sin (180 − ) = x. This suggests that there will be two possible values of
−1
when finding sin
<
<
x=
. 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 θ.
Find the exact value of each expression, if it exists.
73.
ANSWER:
74. sin−1
ANSWER:
75. arctan 1
ANSWER:
76. sin−1
eSolutions Manual - Powered by Cognero
ANSWER:
Page 14
75. arctan 1
ANSWER:
4-7 The Law of Sines and the Law of Cosines
76. sin−1
ANSWER:
Identify the damping factor 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
ANSWER:
f(x) = −2x; the amplitude of the function is decreasing as x approaches 0 from both sides.
78. y =
x cos x
ANSWER:
f(x) =
x; the amplitude of the function is decreasing as x approaches 0 from both sides.
79. y = (x – 1)2 sin x
ANSWER:
2
f(x) = (x – 1) ; the amplitude of the function is decreasing as x approaches 0 from both sides.
80. y = −4x2 cos x
ANSWER:
2
eSolutions
by Cognero
f(x) Manual
= −4x - ;Powered
the amplitude
of
the function is decreasing as x approaches 0 from both sides.
Page 15
4-7 The Law of Sines and the Law of Cosines
80. y = −4x2 cos x
ANSWER:
2
f(x) = −4x ; the amplitude of the function is decreasing as x approaches 0 from both sides.
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.
ANSWER:
Latitude
0
30
45
60
90
Distance
24,881.4
21,547.9
17,593.8
12,440.7
0
The distances range from about 24,881 mi to 0 mi.
82. RADIOACTIVITY A scientist starts with a 1-gram 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.
ANSWER:
a. y = 1.0091(0.9805)x
b. y = 1.0091e−0.0197x
c. about 234 min
Write a polynomial function of least degree with real coefficients in standard form that has the given
zeros.
83. –1, 1, 5
ANSWER:
3
2
Sample answer: f (x) = x – 5x – x + 5
84. –2, −0.5, 4
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ANSWER:
3
Page 16
2
Sample answer: f (x) = 2x – 3x – 18x – 8
83. –1, 1, 5
ANSWER:
3
2
4-7 Sample
The Law
offSines
of Cosines
answer:
(x) = x and
5
– 5x the
– x +Law
84. –2, −0.5, 4
ANSWER:
3
2
Sample answer: f (x) = 2x – 3x – 18x – 8
85. −3, −2i, 2i
ANSWER:
3
2
Sample answer: f (x) = x + 3x + 4x + 12
86. –5i, −i, i, 5i
ANSWER:
4
2
Sample answer: f (x) = x + 26x + 25
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
ANSWER:
E
88. In rDEF, what is the value of θ to the nearest degree?
F 26
G 74
H 80
J 141
ANSWER:
G
89. FREE RESPONSE The pendulum below moves according to θ =
cos 12t, where θ is the angular displacement
in radians and t is the time in seconds.
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Page 17
H 80
J 141
4-7 ANSWER:
The Law of Sines and the Law of Cosines
G
89. FREE RESPONSE The pendulum below moves according to θ =
cos 12t, where θ is the angular displacement
in radians and t is the time in seconds.
a. Graph the function on 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?
ANSWER:
a.
b. The period is
or about 0.524 seconds. This represents the amount of time it takes for the pendulum to complete
one full swing or cycle. The amplitude is
The frequency is
. This represents the maximum angular displacement of the pendulum.
or about 1.91. This represents the number of swings the pendulum completes per second.
c. The maximum angular displacement is 14.3 .
d. The midline represents when the pendulum is vertical and there is no angular displacement.
e . about 0.101 + 0.524n seconds and 0.161 + 0.524n seconds
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