Download Midterm #1A Solutions.

Math 25
February 19, 2017
Name: Answer Key
D. Arnold
Exam #1A
Multiple Choice. (50 points) For each of the following questions, determine the correct
answer then fill in the corresponding oval on your scantron.
(5pts )
1. Convert 5π/12 radians to degrees.
(a) 15◦
(b) 95◦
(d)
4 75◦
(c) 55◦
(e) 45◦
Solution: We know that 180 degrees equals π radians, so:
5π
180 degrees
5π
radians =
radians ·
12
12
π radians
= 75 degrees
(5pts )
2. Convert 36 degrees to radians.
(a) 4π/9 radians
(b) π/8 radians
(d) 3π/5 radians
(c) π/7 radians
(e)
4 π/5 radians
Solution: We know that 180 degrees equals π radians, so:
36 degrees = 36 degrees ·
=
(5pts )
π radians
180 degrees
π
radians
5
3. An arc of a circle subtended by an angle of 80◦ measures 4 inches. What is the radius
of the circle?
(a) 8/π inches
(b)
4 9/π inches
(c) 4/π inches
(d) 3/π inches
(e) 7/π inches
Solution: Consider the following image:
4 inches
80◦
r
Math 25/Exam #1A
– Page 2 of 13 –
Name: Answer Key
The relation between the central angle θ, radius r, and arc length s is given by the
formula:
s
θ=
r
However, in order to use this formula, the angle must be measured in radians. Thus:
π
80◦ = 80◦ ·
180◦
4π
=
9
Next, solve θ = s/r for r:
s
θ=
r
rθ = s
s
r=
θ
Substitute θ = 4π/9 and s = 4 in.
4 in
4π/9
9
r = in
π
r=
(5pts )
4. A sector of a circle subtended by an angle of 75 degrees has an area of 5π/6 square
inches. What is the radius of the sector?
75◦
r
(a) 3/π inches
(b) 4π/3 inches
(c) 2/π inches
(d) 4 inches
(e)
4 2 inches
Solution: Consider the following image:
75◦
r
Math 25/Exam #1A
– Page 3 of 13 –
Name: Answer Key
The area of a sector subtended by an angle θ with radius r is given by the formula
1
A = θr2
2
Let’s solve this equation for r.
2A = θr2
2A
r2 =
θ
r
2A
r=
θ
In order that we may use this formula, our angle θ must be expressed in radian measure.
π
180◦
75◦ = 75◦ ·
=
5π
12
Now, we can substitute θ = 5π/12 and A = 5π/6 square inches.
r
2A
r=
s θ
2
in
2 5π
6
r=
5π
12
p
2
r = 4 in
r = 2 in
(5pts )
5. A young boy ties a tennis ball to the end of a piece of rope that is 36 inches long. He
then swings the ball in a circular arc, the ball traveling about his head at 2 revolutions
per second. What is the linear speed of the ball in inches per second?
(a)
4 144π in/s
(b) 36π in/s
(c) 36 in/s
(d) 72 in/s
(e) 72π in/s
Solution: Start with a picture.
36 in
Math 25/Exam #1A
– Page 4 of 13 –
Name: Answer Key
We will use the formula v = ωr to calculate the velocity of the tennis ball. However, ω
must be measured in radians per unit time.
rev
rad
· 2π
s
s
rad
= 4π
s
ω=2
Next:
v = ωr
rad
(36 in)
= 4π
s
in
= 144π
s
(5pts )
6. Consider the following right triangle.
2
θ
4
Find √
the exact value of csc θ.
(b) 2
(a) 2 5/5
√
(d)
4 5
(e) 1/2
(c)
Solution: Label the hypotenuse h.
h
2
θ
4
Now use the Pythagorean Theorem.
h2 = 22 + 42
h2 = 4 + 16
h2 = 20
√
h = 20
√
h=2 5
√
5/5
Math 25/Exam #1A
– Page 5 of 13 –
Name: Answer Key
Now, the cosecant is the reciprocal of the sine, so let’s first find sin θ.
opp
hyp
2
= √
2 5
1
=√
5
sin θ =
Hence:
csc θ =
(5pts )
√
5
√
2 3
7. Given that sec θ =
, what is the exact value of cos θ?
3
√
√
√
3
6
2
(a)
(b)
(c)
3
√6
√3
3
2
(d)
4
(e)
2
6
Solution: The cosine is the reciprocal of the secant. Hence:
3
cos θ = √
2 3
√
3
3
= √ ·√
2 3
3
√
3 3
=
√6
3
=
2
(5pts )
8. Given the angle θ = 120◦ , which of the following is its reference angle?
(a) 45◦
(b) 5π/6
(c) 30◦
(d)
4 π/3
(e) π/4
Solution: Draw the angle. The reference angle is the angle made with the horizontal
axis, but it should be placed and labeled in the first quadrant.
Math 25/Exam #1A
– Page 6 of 13 –
Name: Answer Key
y
120◦
60◦
x
Hence, the reference angle is 60◦ . However, that choice is not available, so change 60◦
to radians.
πR
60◦ = 60◦ ·
180◦
π
= radians
3
(5pts )
9. What is the period of the function f (x) = 3 tan
(a)
4 4
π x ?
4
(b) 2
(c) 8
(d) 2π
(e) 4π
Solution: The period of y = tan ωx is T = π/ω. Hence:
π
ω
π
= π
4
4π
=
π
=4
T =
(5pts )
10. What is the amplitude of the function f (x) = −2 cos
(a) 2/π
π
2
x−1 ?
(c) −2
(b) 4
(d)
4 2
(e) π/2
Solution: The amplitude of A cos ωx is |A|. Hence:
Amplitude = | − 2|
=2
Math 25/Exam #1A
– Page 7 of 13 –
Name: Answer Key
Essay Questions. (50 points) Do all of your work on graph paper. When finished, arrange
your solutions in order, then staple this page atop your work.
(10pts )
1. A statue sits atop a vertical cliff. You are 500 feet from the base of the cliff. Your angle
of elevation to the bottom of the statue is 52◦ . Your angle of elevation to the top of the
statue is 58◦ . How tall is the statue? Note: First state an exact answer to the problem
(no decimals), then use Mathematica to round your answer to the nearest tenth of a foot.
Solution: First, an image.
x
y
58◦
52◦
500
This image gives us two equations.
tan 52◦ =
y
500
and
tan 58◦ =
x+y
500
Solve the first equation for y.
y = 500 tan 52◦
Substitute this result into the second equation and solve for x.
x+y
500
x
+ 500 tan 52◦
tan 58◦ =
500
tan 58◦ =
Multiply both sides by 500.
500 tan 58◦ = x + 500 tan 52◦
x = 500 tan 58◦ − 500 tan 52◦
Rounding to the nearest tenth of a foot, x = 160.2 feet, the height of the statue.
Math 25/Exam #1A
(10pts )
– Page 8 of 13 –
Name: Answer Key
2. You are given that tan θ = −3 and π/2 < θ < π. Set up an xy-coordinate system, draw
and label the angle θ, then choose and label an appropriate point on the terminal side of
the angle θ. Use this information to compute the five remaining trigonometric functions
of θ. Make sure each answer is in simple radical form. Note: No right triangles may be
drawn for this problem.
Solution: Start with an image.
(−1, 3)
√
y
10
θ
x
Hence:
√
−1
x
10
cos θ = = √ = −
r
10
10
√
y
3
3 10
sin θ = = √ =
r
10
10
y
3
=
= −3
x
−1
x
−1
1
cot θ = =
=−
y
3
3
√
√
r
10
sec θ = =
= − 10
x
−1
√
√
r
10
10
csc θ = =
=
y
3
3
tan θ =
(5pts
ea. )
3. Sketch exactly one period of each of the following functions on graph paper. Clearly
scale and label each axis. Each of these problems must be sketched by hand without the
aid of a technology.
1
(a) y = − tan
x
2
Math 25/Exam #1A
– Page 9 of 13 –
Solution: The period of y = − tan
1
x
2
Name: Answer Key
is:
π
1/2
= 2π
T =
We will start by drawing the graph of y = tan
1
x
2
.
y
π
Now, to draw the graph of y = − tan
the x-axis.
1
x
2
2π
x
, we need only flip the last graph across
y
π
π (b) y = 2 sec
x
2
Solution: To start, the period of y = 2 sec
T =
2π
π
x
2
x
is:
2π
π
2
=4
We’ll start by drawing the graph of y = sec
π
x
2
. Because the secant is the inverse
Math 25/Exam #1A
– Page 10 of 13 –
Name: Answer Key
of the cosine, we lightly draw the cosine curve, then the secant will have vertical
asymptotes where the cosine is equal to zero.
y
1
2
4
x
−1
Now, to draw the graph of y = 2 sec
π
x
2
, we need only double all the y-values.
y
2
2
4
x
−2
(c) y = −2 sin(2x − π)
Solution: Start by factoring out a 2.
y = −2 sin(2x − π)
π
y = −2 sin 2 x −
2
We’ll begin by drawing the graph of y = −2 sin 2x, which has an amplitude of 2, a
period of T = 2π/2 = π, and is reflected across the x-axis.
Math 25/Exam #1A
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Name: Answer Key
y
2
π/2
x
π
−2
Next, replace x in y = −2 sin 2x with x − π2 to obtain y = −2 sin 2 x −
will shift our previous graph π/2 units to the right.
π
2
, which
y
2
π/2
π
x
3π/2
−2
1
π
(d) y = 3 cos
x+
2
4
Solution: Start by factoring out a 1/2.
1
π
y = 3 cos
x+
2
4
1
π
y = 3 cos
x+
2
2
We’ll begin by drawing the graph of y = 3 cos 21 x, which has an amplitude of 3 and
a period of T = 2π/(1/2) = 4π.
y
3
π
2π
3π
4π
x
−3
Next, replace x in y = 3 cos 12 x with x + π2 to obtain y = 3 cos 12 x +
shift our previous graph π/2 units to the left.
π
2
, which will
Math 25/Exam #1A
– Page 12 of 13 –
Name: Answer Key
y
3
x
−π/2
π/2
3π/2
5π/2
7π/2
−3
(10pts )
4. Pictured below is one period of a trigonometric function. Determine its equation.
y
2
π
x
π/2
−3
Solution: This picture appears to be a cosine function with period T = π, amplitude
2, that has been reflected across the x-axis, then shifted downward. So, let’s begin by
graphing y = −2 cos 2x, which has period T = 2π/2 = π.
Math 25/Exam #1A
– Page 13 of 13 –
Name: Answer Key
y
2
π
x
π/2
−2
Next, subtract 1 from the equation y = −2 cos 2x, which gives us the equation y =
−2 cos 2x − 1 and shifts the previous graph down 1 unit.
y
2
π
x
π/2
−3
That is the given picture. Hence, we have the correct equation, y = −2 cos 2x − 1.