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Exam 3
Math 1151, Precalculus 2
Name:
Summer 2012
Instructions.
1. Show all the steps of your work clearly.
2. No graphing calculators, iPods, phones or computers
Question Points Your Score
Q1
30
Q2
20
Q3
25
Q4
30
Q5
10
Q6
15
Q7
30
Q8
20
TOTAL
180
Q1]. . . [30 points]
Find the area of the shape below. Be sure to write down all the formulas you use.
The larger triangle has an area of 21 (50)(100) sin(40◦ ) = 1606.97. To find the area the smaller triangle
we first find the length of the side that the triangles share using the larger triangle and the law of cosines.
p
a = 502 + 1002 − 2(50)(100) cos(40◦ ) = 69.57.
Then we used the law of sines to find the angle opposite the side of length 20:
◦
)
sin−1 (20 sin(100
) = 16.45◦ . Then B = 180◦ − 100◦ − 16.45◦ = 63.55◦ . Now we can find the area of
69.57
the smaller triangle which is 21 (20)(69.57) sin(63.55◦ ) = 622.88. Therefore the area of the whole shape is
1606.97+622.88=2229.9.
Q2]. . . [20 points]
Establish the identity
cot(θ) − tan(θ)
= cos(2θ)
cot(θ) + tan(θ)
cot(θ) − tan(θ)
=
cot(θ) + tan(θ)
=
=
cos θ
sin θ
− cos
sin θ
θ
cos θ
sin θ
+ cos θ
sin θ
cos2 θ−sin2 θ
sin θ cos θ
cos2 θ+sin2 θ
sin θ cos θ
2
2
cos θ − sin θ
cos2 θ + sin2 θ
= cos(2θ).
Q3]. . . [25 points]
Joe is at point Q which is 400 feet from an office building. Joe wants to measure the length of the windows
on the top floor of the office building, so he measures the angle of elevation to the base of the window to
be 64.5◦ and the angle of elevation to the top of the window is 65◦ . What is the length of the window?
(round to the nearest foot) Be sure to label the figure.
There are two right triangles here and the quantity we want is b0 − b = 400 tan(65◦ ) − 400 tan(64.5◦ ) =
857.8 − 838.6 = 19.2 ft.
Q4]. . . [30 points]
Consider a triangle with side lengths b = 2, c = 3 and angle B = 40◦ . Determine if any triangles arise
from the given information. If any triangles do result solve all such triangles. Be sure to write down all
formulas you use and round to the nearest tenth.
This is a SSA triangle which is the ambiguous case. Using the Law of sines we find that
sin(40◦ )
2
=
sin C
3
so that C1 = 74.6◦ or C2 = 180 − C1 = 105.4◦ . So two triangles arise because C2 + B < 180◦ .
◦
)
Triangle 1: A1 = 180 − B − C1 = 65.4◦ . The law of sines gives a1 = 2 sin(65.4
= 2.8.
sin(40◦ )
◦
)
Triangle 2: A2 = 180 − B − C2 = 34.6◦ . The law of sines gives a2 = 2 sin(34.6
= 1.8
sin(40◦ )
Q5]. . . [10 points]
).
Let P be the point with polar coordinates (2, − 2π
3
(a) Plot the point P .
√
(b) What are the rectangular coordinates of the point P ? (x, y) = (2 cos(−2π/3), 2 sin(2π/3)) = (−1, − 3).
Q6]. . . [15 points]
Find other polar coordinates (r, θ) for the point P above for which
(a) r < 0 and 0 ≤ θ < 2π (−2, −2π
+ π) = (−2, π3 )
3
(b) r > 0 and 2π ≤ θ < 4π (2, −2π
+ 4π) = (2, 10π
)
3
3
Q7]. . . [30 points]
A boat leaves point B bound for point C a distance of 500 miles away. After traveling for 6 hours at a
constant speed of 25 miles per hour the captain realizes that he is off course by 13◦ , see the illustration.
Write down all formulas you use and round to the nearest tenth.
(a) How far is the boat from C when they realize the mistake?
(b) Through what angle should the boat turn to travel directly to point C?
(a) (25)(6)=150 mi
(b) By the Law of Cosines b =
p
5002 + 1502 − 2(150)(500) cos(13◦ ) = 355.4. Then using the Law of
2
2
2
−355.4 −150
cosines again we can find the angle A = cos−1 ( 500−2(355.4)(150)
) = 161.6◦ and so the boat must turn through
an angle of 180 − A = 18.4◦ to go directly to C.
Q8]. . . [20 points]
Solve the following trigonometric equation on the interval [0, 2π).
tan(2θ) + 2 cos(θ) = 0
tan(2θ) + 2 cos(θ) =
sin(2θ)
+ 2 cos θ
cos(2θ)
If cos(2θ) 6= 0 we can multiply both sides by cos(2θ) and concentrate on the numerator
0 = sin(2θ) + 2 cos(2θ) cos θ
= 2 sin θ cos θ + 2(1 − 2 sin2 θ) cos θ
= −2 cos θ(2 sin2 θ − sin θ − 1)
= −2 cos θ(2 sin θ + 1)(sin θ − 1)
So now we just have to solve for when the three factors are equal to 0.
• cos θ = 0 when θ is π2 , 3π
2
• sin θ = − 12 when θ is
7π 11π
, 6
6
• sin θ = 1 when θ is π2 .
And note that cos(2θ) 6= 0 for any of these values. Thus the solution set is { π2 , 3π
, 7π
, 11π
}.
2
6
6