# Practice Test - Chapter 4

```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: 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.
4. SOLUTION: Because the length of the side opposite and the hypotenuse are given, use the cosine function.
5. MULTIPLE CHOICE What is the linear speed Find the measure of angle θ. Round to the
nearest degree, if necessary.
3. SOLUTION: Because the length of the side opposite and adjacent
to θ are given, use the tangent function.
of a point rotating at an angular speed of 36 radians
per second at a distance of 12 inches from the
center of the rotation?
A 420 in./s
B 432 in./s
C 439 in./s
D 444 in./s
SOLUTION: The formula for linear speed is v =
where s is the
arc length traveled. Arc length is equal to rθ, where
r is the radius and θ is the central angle.
Thus,
. The distance from the center of
rotation, also known as the radius, is 12 inches,
so
.
Page 1
The correct choice is B.
Practice Test - Chapter 4
5. MULTIPLE CHOICE What is the linear speed of a point rotating at an angular speed of 36 radians
per second at a distance of 12 inches from the
center of the rotation?
A 420 in./s
B 432 in./s
C 439 in./s
D 444 in./s
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
SOLUTION: The formula for linear speed is v =
where s is the
arc length traveled. Arc length is equal to rθ, where
r is the radius and θ is the central angle.
Thus,
. The distance from the center of
rotation, also known as the radius, is 12 inches,
so
.
7. SOLUTION: To convert a radian measure to degrees, multiply by
The formula for angular speed is
speed is 36 radians per second, so
. The angular
.
Therefore,
8. Find the area of the sector of the circle shown.
The correct choice is B.
Write each degree measure in radians as a
multiple of π and each radian measure in
degrees.
6. 200
SOLUTION: SOLUTION: The area of a sector A =
r
2
, where r is the
radius and θ is the central angle.
To convert a degree measure to radians, multiply by
7. Page 2
The area of the sector is about 209.4 square inches.
Practice Test - Chapter 4
8. Find the area of the sector of the circle shown.
10. SOLUTION: The terminal side of
lies in Quadrant IV. Therefore, its reference angle is
SOLUTION: The area of a sector A =
r
2
' =
.
, where r is the
radius and θ is the central angle.
Find the exact value of each expression.
The area of the sector is about 209.4 square inches.
Sketch each angle. Then find its reference
angle.
9. 165
11. sec
SOLUTION: Because the terminal side of θ lies in Quadrant III,
the reference angle θ&acute; of SOLUTION: is .
The terminal side of 240 lies in Quadrant II.
Therefore, its reference angle is ' = 180 –165 or 15 .
10. SOLUTION: The terminal side of
lies in Quadrant IV. Therefore, its reference angle is
.
' =
12. cos (−240 )
SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 )
Because the terminal side of lies in Quadrant II,
the reference angle ' is .
Page 3
this additional restriction, must lie in the 3rd
quadrant. With sin &lt; 0 and cos θ &lt; 0, cot must be &gt; 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.
12. cos (−240 )
SOLUTION: cos (−240 ) =cos (−240 +360) = cos (120 )
Because the terminal side of lies in Quadrant II,
the reference angle ' is .
14. y = 4 cos
– 5
SOLUTION: In this function, a = 4, b =
, c = 0, and d = –5.
13. MULTIPLE CHOICE An angle satisfies the following inequalities: csc θ &lt; 0, cot θ &gt; 0, and sec θ
&lt; 0. In which quadrant does lie?
FI
G II
H III
J IV
Graph y = 4 cos
shifted 5 units down.
SOLUTION: If csc &lt; 0, then sin &lt; 0. So, must lie in the 3rd
or 4th quadrant. If sec &lt; 0, then cos &lt; 0. With this additional restriction, must lie in the 3rd
quadrant. With sin &lt; 0 and cos θ &lt; 0, cot must be &gt; 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: In this function, a = 4, b =
15. , c = 0, and d = –5.
SOLUTION: In this function, a = –1, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
Page 4
Practice Test - Chapter 4
16. TIDES The table gives the approximate times that
15. the high and low tides occurred in San Azalea Bay
over a 2-day period.
SOLUTION: In this function, a = –1, b = 1, c = , and d = 0.
Because d = 0, there is no vertical shift.
a. The tides can be modeled with a trigonometric
function. Approximately what is the period of this
function?
b. The difference in height between the high and low
tides is 7 feet. What is the amplitude of this function?
c. Write a function that models the tides where t is
measured in hours. Assume the function has no
phase shift or vertical shift.
SOLUTION: a. The period lasts from peak to peak, or from high
tide to high tide. 3:04 PM is 12 hours and 30 minutes
after 2:35 AM.
b. The amplitude is one half of the difference
between the maximum (high tide) and the minimum
(low tide).
One half of 7 is 3.5.
c. The period is 12 hours and 30 minutes, or 12.5
hours.
Graph y = – sin x shifted
units to the left.
16. TIDES The table gives the approximate times that
the high and low tides occurred in San Azalea Bay
over a 2-day period.
a. The tides can be modeled with a trigonometric
function. Approximately what is the period of this
function?
b. The difference in height between the high and low
tides is 7 feet. What is the amplitude of this function?
c. Write a function that models the tides where t is
measured in hours. Assume the function has no
phase shift or vertical shift.
SOLUTION: a. The period lasts from peak to peak, or from high
tide to high tide. 3:04 PM is 12 hours and 30 minutes
after 2:35 AM.
eSolutions
b. The amplitude is one half of the difference
between the maximum (high tide) and the minimum
Locate the vertical asymptotes, and sketch the
graph of each function.
17. SOLUTION: The graph of
x translated
is the graph of y = tan
units to the left. The period is . Find the location of two consecutive vertical
asymptotes.
Page
or 5
17. SOLUTION: 18. y =
sec (2x)
Practice Test - Chapter 4
is the graph of y = tan
The graph of
x translated
units to the left. The period is or . Find the location of two consecutive vertical
asymptotes.
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.
and and Create a table listing the coordinates of key points
for
for one period on
.
Create a table listing the coordinates of key points
Function
Vertical
Asymptote
Intermediate
Point
x-int
.
Function
Vertical
Asymptote
Intermediate
Point
x-int
(0, 0)
Intermediate
Point
Vertical
Asymptote
(0, 1)
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
18. y =
sec 2x for one period on
for y =
y = tan x
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.
sec (2x)
The graph of y =
sec 2x is the graph of y = sec x
compressed horizontally and compressed vertically.
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.Page 6
19. a = 8, b = 16, A = 22
Practice
Test - Chapter 4
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
When B 131 , then C 180 – (22 + 131 )
or about 27 . Apply the Law of Sines to find c.
SOLUTION: Draw a diagram of a triangle with the given
dimensions.
Notice that A is acute and a &lt; b because 8&lt; 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: Draw a diagram of a triangle with the given
dimensions.
8 &gt; 6, so two solutions exist.
Apply the Law of Sines to find B.
Notice that A is acute and a &gt; b because 9 &gt; 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.
eSolutions
Manual
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When
B - 131
, then
C 180
Page 7
– (22 + 131 )
or about 27 . Apply the Law of Sines to find c.
Therefore, the remaining measures of
are
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 &gt; b because 9 &gt; 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.
Use the Law of Sines to find a missing angle
Use the Law of Sines to find a missing angle
measure.
Page 8
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 =
.
24. –1
. Therefore, cos
25. NAVIGATION A boat leaves a dock and travels 45&ordm; north of west averaging 30 knots for 2 hours.
The boat then travels directly west averaging 40
knots for 3 hours.
Page 9
25. NAVIGATION A boat leaves a dock and travels 45&ordm; north
of west
averaging 30
Practice
Test
- Chapter
4 knots for 2 hours.
The boat then travels directly west averaging 40
knots for 3 hours.
a. How many nautical miles is the boat from the
dock after 5 hours?
b. How many degrees south of east is the dock from
the boat’s present position?
SOLUTION: a. During the first leg of the trip the boat traveled 30
knots for 2 hours, and therefore traveled a distance
of 30 &middot; 2 or 60 nautical miles. During the second leg of the trip, the boat traveled 40 knots for 3 hours, so
the distance the boat traveled is 40 &middot; 3 or 120 nautical miles. Draw a diagram to model the situation. Let x
represent the distance the boat has traveled from the
dock after 5 hours.
Use the Law of Cosines to find b.
Therefore, the boat is 167.9 nautical miles from the
dock after 5 hours.
b. Use the Law of Sines to find C.
Therefore, the dock is about 15 south of east from
the boat's current position.