Practice Test - Chapter 4
Find the value of x. Round to the nearest tenth,
if necessary.
Find the measure of angle θ. Round to the
nearest degree, if necessary.
1. SOLUTION: 3. SOLUTION: An acute angle measure and the length of the
hypotenuse are given, so the sine function can be
used to find the length of the side opposite .
Because the length of the side opposite and adjacent
to θ are given, use the tangent function.
2. SOLUTION: 4. An acute angle measure and the length of the side
opposite the angle are given, so the tangent function
can be used to find the length of the side adjacent to
the angle.
Find the measure of angle θ. Round to the
nearest degree, if necessary.
SOLUTION: Because the length of the side opposite and the hypotenuse are given, use the cosine function.
Write each degree measure in radians as a
multiple of π and each radian measure in
degrees.
6. 200
SOLUTION: To convert a degree measure to radians, multiply by
3. SOLUTION: Because the length of the side opposite and adjacent
to θ are given, use the tangent function.
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7. Page 1
SOLUTION: Practice Test - Chapter 4
The area of the sector is about 209.4 square inches.
Sketch each angle. Then find its reference
angle.
9. 165
7. SOLUTION: SOLUTION: To convert a radian measure to degrees, multiply by
The terminal side of 240 lies in Quadrant II.
Therefore, its reference angle is ' = 180 –165 or 15 .
8. Find the area of the sector of the circle shown.
10. SOLUTION: The terminal side of
SOLUTION: The area of a sector A =
r
2
, where r is the
Therefore, its reference angle is
Sketch each angle. Then find its reference
angle.
9. 165
' =
.
radius and θ is the central angle.
The area of the sector is about 209.4 square inches.
lies in Quadrant IV. Find the exact value of each expression.
11. sec
SOLUTION: Because the terminal side of θ lies in Quadrant III,
SOLUTION: The terminal side of 240 lies in Quadrant II.
Therefore, its reference angle is ' = 180 –165 or 15 .
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the reference angle θ´ of is .
Page 2
Practice Test - Chapter 4
Find the exact value of each expression.
11. sec
SOLUTION: Because the terminal side of θ lies in Quadrant III,
the reference angle θ´ of is .
13. MULTIPLE CHOICE An angle satisfies the following inequalities: csc θ < 0, cot θ > 0, and sec θ
< 0. In which quadrant does lie?
FI
G II
H III
J IV
SOLUTION: If csc < 0, then sin < 0. So, must lie in the 3rd
or 4th quadrant. If sec < 0, then cos < 0. With this additional restriction, must lie in the 3rd
quadrant. With sin < 0 and cos θ < 0, cot must be > 0.
The correct choice is H.
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 measurements to the nearest degree.
19. a = 8, b = 16, A = 22
SOLUTION: 12. cos (−240 )
Draw a diagram of a triangle with the given
dimensions.
SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 )
Because the terminal side of lies in Quadrant II,
the reference angle ' is .
Notice that A is acute and a < b because 8< 16.
Therefore, two solutions may exist. Find h.
13. MULTIPLE CHOICE An angle satisfies the following inequalities: csc θ < 0, cot θ > 0, and sec θ
< 0. In which quadrant does lie?
FI
G II
H III
J IV
SOLUTION: eSolutions Manual - Powered by Cognero
If csc < 0, then sin < 0. So, must lie in the 3rd
or 4th quadrant. If sec < 0, then cos < 0. With 8 > 6, so two solutions exist.
Apply the Law of Sines to find B.
Page 3
Therefore, the remaining measures of
B 49 , C 109 , c 20.1 and B
27 , c 9.7.
8 > 6, so two solutions exist.
Practice
Test - Chapter 4
are
131 , C
Apply the Law of Sines to find B.
20. a = 9, b = 7, A = 84
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Because two angles are now known, C 180 –
(22 + 49 ) or about 109 . Apply the Law of Sines
to find c.
Notice that A is acute and a > b because 9 > 7.
Therefore, one solution exists. Apply the Law of
Sines to find B.
When B 131 , then C 180 – (22 + 131 )
or about 27 . Apply the Law of Sines to find c.
Because two angles are now known, C ≈ 180 –
(84 + 51 ) or about 45 . Apply the Law of Sines
to find c.
Therefore, the remaining measures of
B 49 , C 109 , c 20.1 and B
27 , c 9.7.
are
131 , C
20. a = 9, b = 7, A = 84
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Therefore, the remaining measures of
B 51 , C 45 , and c 6.4.
are
21. a = 3, b = 5, c = 7
SOLUTION: Use the Law of Cosines to find an angle measure.
Notice that A is acute and a > b because 9 > 7.
Apply the Law of
Sines to find B.
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Therefore,
solution
exists.
Page 4
Use the Law of Sines to find a missing angle
Find the measure of the remaining angle.
Therefore,
the -remaining
measures
of
Practice
Test
Chapter
4
B
51 , C
45 , and c
are
6.4.
Therefore, A
22 , B
38 , and C
120 .
22. a = 8, b = 10, C = 46
21. a = 3, b = 5, c = 7
SOLUTION: SOLUTION: Use the Law of Cosines to find an 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, A
22 , B
38 , and C
Therefore, c
7.3, A
52 , and B
82 .
120 .
22. a = 8, b = 10, C = 46
SOLUTION: Use the Law of Cosines to find the missing side
measure.
Use the Law of Sines to find a missing angle
measure.
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