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Math 142 - Midterm 2 Version 2 Solutions
1. (10 points) You are piloting a ship from A to C. Shallow water crosses the direct route, so
you must first go to point B to avoid the shallows. You are currently 12 miles from C and 5
miles from B. The difference in headings to B and C is 20◦ .
(a) (5 points) How far is it from B to C?
Call the unknown right-hand side a. Writing the law of cosines using the angle at B,
we get
a2 = 122 + 52 − 2 · 12 · 5 · cos(20◦ )
≈ 56.2369
a ≈ 7.499 mi.
(b) (5 points) When you reach B, how far must you turn (i.e. by how many degrees must
you change your course)?
The angle we want is θ in the diagram above, which is 180◦ − β. So let’s find β using
the law of cosines:
122 = 52 + 7.4992 − 2 · 5 · 7.499 · cos(β)
122 − 52 − 7.4992
−2 · 5 · 7.499
≈ −0.83695
cos(β) =
β ≈ 146.82◦ .
So θ = 180 − 146.82 = 33.18◦ .
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2. (10 points) You are laying an underwater cable from island D to island C. A surveying crew
has provided you with the following measurements: 6 BAD = 105◦ , 6 ABD = 25◦ , 6 DBC =
30◦ , 6 BCD = 80◦ . Also, you know that the distance along the shore from point A to point
B is 10 miles. How much cable do you need?
In the bottom triangle, the top angle is 180 − 105 − 25 = 50◦ . Now we can use the law of
sines in the bottom triangle to find the side I’ve labelled x:
sin(50◦ )
sin(105◦ )
=
10
x
sin(105◦ )
· 10
x=
sin(50◦ )
≈ 12.6092661 mi.
Now that we have x, we can use the law of sines on the top triangle to find the side from D
to C, which I’ve labelled y:
sin(80◦ )
sin(30◦ )
=
12.6092661
y
sin(30◦ )
y=
· 12.6092661
sin(80◦ )
≈ 6.40189 mi.
So we need 6.40189 miles of cable.
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3. (10 points) Find exact expressions for the following. (Decimal approximations are not sufficient for full credit.)
(a) (5 points) If sin(θ) = 1/3 with π/2 ≤ θ ≤ π and cos(φ) = 1/7 with −π/2 ≤ φ ≤ 0, find
cos(θ + φ).
We’ll use the addition formula for cosine, so we’ll need to know cos(θ) and sin(φ).
q
√
cos(θ) = ± 1 − sin2 (θ) = ± 38 .
√
But θ is in the second quadrant, so cos(θ) is negative, so cos(θ) = −
8
3 .
Next,
√
p
sin(φ) = ± 1 − cos2 (φ) = ± 748 .
√
Since φ is in the fourth quadrant, sin(φ) is negative, so sin(φ) = −
to use the sum formula for cosine:
48
7 .
cos(θ + φ) = cos(θ) cos(φ) − sin(θ) sin(φ)
√
8 1
3 )( 7 )
√ √
− 8+ 48
.
21
= (−
=
(b) (5 points) cos( 7π
12 ) = cos
7π
6
− ( 13 )(−
√
48
7 )
!
2
We can use the half angle identity for cosine:
q
1
7π
cos( 7π
)
=
±
12
2 (1 + cos( 6 ))
q
√
= ± 12 − 43 .
But
7π
12
is in the second quadrant, so the cosine must be negative. So
cos( 7π
12 )
q
= − 12 −
√
3
4 .
Now we’re ready
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4. (a) (5 points) Write the complex number 1 −
i sin(θ))).
√
3i in polar form (i.e. in the form r(cos(θ) +
We need to find r and θ, then put the in the polar form.
q
√
r = (1)2 + 3)2 = 2.
Note that since we’ve chosen the positive value for r (which we usually do when writing
complex numbers in polar form), we have to be careful to pick the right θ value. We
have
√
√
x
− 3
tan(θ) = =
= − 3.
y
1
This is one of the common values of tangent
√ for which we know θ. In fact, θ is either
−π/3 or 2π/3. The complex number 1 − 3i is in the fourth quadrant of the complex
plane, so with positive r, we must pick θ = −π/3. Now we know r and θ, so
√
1 − 3i = 2(cos(− π3 ) + i sin(− π3 )).
(b) (5 points) Find the polar form of the complex number that results from the complex
product
√
(1 − 3i) · 3(cos(4) + i sin(4)).
For full credit, your answer must be exact. A decimal approximation is not sufficient
for full credit. (Hint: use your result from part (a).)
Remember that complex multiplication behaves nicely when you write the complex
numbers in polar form:
r coordinates multiply and θ coordinates add. So, using the
√
polar form for 1 − 3i that we found in (a),
√
2π
(1 − 3i) · 3(cos(4) + i sin(4)) = 2(cos(− 2π
3 ) + i sin(− 3 )) · 3(cos(4) + i sin(4))
2π
= 2 · 3 · (cos(− 2π
3 + 4) + i sin(− 3 + 4))
2π
= 6(cos(− 2π
3 + 4) + i sin(− 3 + 4)).
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5. (a) (5 points) Write all choices of polar coordinates for the point (x, y) = (−1, −1) such
that −2π ≤ θ ≤ 2π.
We have
p
p
√
r = ± x2 + y 2 = ± (−1)2 + (−1)2 = ± 2.
Also,
y
−1
=
= 1.
x
−1
So θ = π4 + πk. We have to be careful to get the right possibilities for θ with the right
possibilities for r so we end up in the right quadrant (the 3rd quadrant, in the case of
(−1, −1)c . We get
√
r= 2
θ = 5π
or − 3π
4
4 ,
√
r=− 2
θ = π4 or − 7π
4 .
tan θ =
(There are other possible values for θ, but they lie outside the given interval −2π ≤ θ ≤
2π.)
(b) (5 points) Convert the equation (x − 2)2 + (y − 3)2 = 13 into a polar equation. Simplify
the result as much as you can.
Eventually, we’ll replace x with r cos θ and y with r sin θ, but first let’s do some algebra:
(x − 2)2 + (y − 3)2 = 13
x2 − 4x + 4 + y 2 − 6y + 9 = 13
x2 + y 2 − 4x − 6y = 0
r2 − 4r cos θ − 6r sin θ = 0
r − 4 cos θ − 6 sin θ = 0
since r2 = x2 + y 2
(I multiplied by 1/r to get this)
r = 4 cos θ + 6 sin θ.