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.
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
14. y = 4 cos
– 5
SOLUTION: 12. cos (−240 )
In this function, a = 4, b =
, c = 0, and d = –5.
SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 )
Because the terminal side of lies in Quadrant II,
the reference angle ' is .
Graph y = 4 cos
shifted 5 units down.
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 Page 3
this additional restriction, must lie in the 3rd
quadrant. With sin < 0 and cos θ < 0, cot must be > 0.
Practice
Test - Chapter 4
The correct choice is H.
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
14. y = 4 cos
15. SOLUTION: – 5
In this function, a = –1, b = 1, c = SOLUTION: In this function, a = 4, b =
, and d = 0.
Because d = 0, there is no vertical shift.
, c = 0, and d = –5.
Graph y = – sin x shifted
units to the left.
Graph y = 4 cos
shifted 5 units down.
Locate the vertical asymptotes, and sketch the
graph of each function.
17. SOLUTION: is the graph of y = tan
The graph of
x translated
15. or . Find the location of two consecutive vertical
asymptotes.
SOLUTION: In this function, a = –1, b = 1, c = units to the left. The period is , and d = 0.
and Because d = 0, there is no vertical shift.
Create a table listing the coordinates of key points
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for
for one period on
.
sec 2x is the graph of y = sec x
The graph of y =
and compressed horizontally and compressed vertically.
The period is
Practice Test - Chapter 4
Create a table listing the coordinates of key points
for
for one period on
or π. Find the location of two
vertical asymptotes.
.
and Function
y = tan x
Vertical
Asymptote
Intermediate
Point
x-int
Create a table listing the coordinates of key points
sec 2x for one period on
for y =
(0, 0)
Intermediate
Point
Vertical
Asymptote
.
Function
(0, 1)
Vertical
Asymptote
Intermediate
Point
x-int
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
y = sec x
y=
sec 2x
(0, 1)
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
18. y =
sec (2x)
SOLUTION: sec 2x is the graph of y = sec x
The graph of y =
compressed horizontally and compressed vertically.
The period is
or π. Find the location of two
vertical asymptotes.
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: and eSolutions Manual - Powered by Cognero
Create a table listing the coordinates of key points
Draw a diagram of a triangle with the given
dimensions.
Page 5
angle measurements to the nearest degree.
19. a = 8, b = 16, A = 22
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
When B 131 , then C 180 – (22 + 131 )
or about 27 . Apply the Law of Sines to find c.
Practice Test - Chapter 4
Notice that A is acute and a < b because 8< 16.
Therefore, two solutions may exist. Find h.
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: 8 > 6, so two solutions exist.
Draw a diagram of a triangle with the given
dimensions.
Apply the Law of Sines to find B.
Notice that A is acute and a > b because 9 > 7.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Because two angles are now known, C 180 –
(22 + 49 ) or about 109 . 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.
When B 131 , then C 180 – (22 + 131 )
or about 27 . Apply the Law of Sines to find c.
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Therefore, the remaining measures of
B 51 , C 45 , and c 6.4.
arePage 6
Therefore, the remaining measures of
B 49 , C 109 , c 20.1 and B
Practice
27 , Test
c 9.7.- Chapter 4
are
131 , C
Therefore, the remaining measures of
B 51 , C 45 , and c 6.4.
are
20. a = 9, b = 7, A = 84
21. a = 3, b = 5, c = 7
SOLUTION: SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Use the Law of Cosines to find an angle measure.
Notice that A is acute and a > b because 9 > 7.
Therefore, one solution exists. Apply the Law of
Sines to find B.
Use the Law of Sines to find a missing angle
measure.
Because two angles are now known, C ≈ 180 –
(84 + 51 ) or about 45 . Apply the Law of Sines
to find c.
Find the measure of the remaining angle.
Therefore, A
22 , B
38 , and C
120 .
22. a = 8, b = 10, C = 46
SOLUTION: Use the Law of Cosines to find the missing side
measure.
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.
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Use the Law of Sines to find a missing angle
Use the Law of Sines to find a missing angle
measure.
Page 7
Find the measure of the remaining angle.
Practice Test - Chapter 4
22 , B
Therefore, A
38 , and C
Find the measure of the remaining angle.
120 .
7.3, A
Therefore, c
52 , and B
82 .
Find the exact value of each expression, if it
exists.
22. a = 8, b = 10, C = 46
SOLUTION: Use the Law of Cosines to find the missing side
measure.
23. SOLUTION: Find a point on the unit circle on the interval
with a x-coordinate of
.
Use the Law of Sines to find a missing angle
measure.
When t =
=
, cos t =
–1
. Therefore, cos
.
Find the measure of the remaining angle.
7.3, A
Therefore, c
52 , and B
82 .
Find the exact value of each expression, if it
exists.
24. SOLUTION: Find a point on the unit circle on the interval
23. with a y-coordinate of
.
SOLUTION: Find a point on the unit circle on the interval
with a x-coordinate of
.
When t =
, sin t =
. Therefore, sin
–1
=
.
When t =
=
, cos t =
.
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24. –1
. Therefore, cos
Page 8