MAT187 Pre-Calculus
Exam 3A, Hints and Answers
1) How many revolutions will a car wheel of diameter 30 inches make as a car travels
one mile? One mile is 5280 feet.
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n  number of revolutions. s  distance traveled = n(circumference).
5280(12)  n(30 ) , n = 672.27 revolutions.
2) To estimate the height of a mountain above a level plain, the angle to the top of the
mountain is measured to be 32 degrees. One thousand feet closer to the mountain on
the horizontal plain, the angle is measured to be 35 degrees. Estimate the height of
the mountain.
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tan 32 
h
h
, tan 35 
1000  x
x
mountain top
Solving this algebraically gives
h  5805 ft
h
32°
1000 ft
35°
x
3) Without using your calculator, except to check your answer, find the exact value of
11
sin
.
6
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11

 1
11
1
 2  , which is in QIV. sin  , so sin
 .
6
6
6 2
6
2
4) A triangle with sides a, b, c, and opposite angles A, B, C, has b = 73, c = 82, and
angle B = 58 degrees. Solve all possible triangles that satisfy the given conditions.
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Two solutions (see figure):
73
82

.
sin C1
sin 58
C1  72.29 , C2  180  C1  107.71 , A1  79.71 .
5) A boy is flying two kites at the same time. The line to one of the kites is 380 feet
long, and the line to the other kite is 420 feet long. He estimates the angle between
the two lines to be 30 degrees. Approximate the distance between the two kites.
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Law of Cosines:
420 feet
x
30°
380 feet
x 2  4202  3802  2(420)(380) cos30
x  211 feet
6) Solve cos 2x  cos x  2 in the interval [0, 2 ) without using your calculator,
except to check your answer.
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cos 2 x  2 cos 2 x  1 , so given equation becomes 2 cos 2 x  1  cos x  2 , or
2 cos 2 x  cos x  3  0 , which factors into (2 cos x  3)(cos x  1)  0 , and, cosx=1,
3
so x = 0. The other factor produces x   , which is impossible.
2
7) Find the exact value of tan15 without using your calculator, except to check your
answer.
_________________________________
30
Two ways: use the tangent half angle formula ( tan
), or the difference formula
2
tan(45  30 ) to get the result 2  3 .
8) Rewrite cos(2 tan 1 x) as an algebraic function of x.
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tan 1 x   . cos 2  2 cos 2   1
2
 1 
, cos 2  2 
cos  
  1
 2
x

1
x2  1


2
1 x
.
cos 2 
1  x2
1
x2+1
x
θ
1
9) A 380 feet tall building supports a 40 feet tall communications tower. As a driver
approaches the building the viewing angle of the tower (top to bottom) changes. At
what distance from the building is the viewing angle as large as possible?
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40 ft
380 ft
  tan 1
380  40
380
 tan 1
.
x
x
Use calculator to see that max occurs at
about 400 feet.
θ
x
10) Let x   tan  ,

 

. Simplify
x
2
2
 2  x2
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tan  sin 

cos   sin  ,
 x
 1  tan  sec cos 
cos is not zero in the given domain.
y
x
2
2

 tan 
.
2

Find the exact value (no decimals) of cos 225 without using your calculator except to
to check your answer.
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cos 225   cos(45 )  
2
.
2
1
12) If cos    , and  is in QIII, find tan  cot   csc .
3
_________________________________
1
2

2 . QIII makes it minus. Thus
9
3
3
2
2
2, cot  
sin   
2 . Continue to get tan   2 2, csc   
.
4
3
43 2
Combine and simplify to get tan  cot   csc  
.
4
sin    1  cos 2    1 
It’s easier to observe that the first term is just 1, so we have only to deal with
1  csc , and you’ll get the same answer.
13) Find the area of the shaded region
shaded
π
3
shaded
12
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1

3
6 3.
The area of the triangle = (12)h, h  12sin  12
2
3
2
1

So area of triangle = 36 3 . Area of the sector is (122 )  24 . Thus the shaded
2
3
2
area is  (12 )  24  36 3  120  36 3  439.3 square units.
14) Find the exact value (no decimals) of csc(cos 1
7
) without using your calculator
25
except to check your answer.
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15) Find the exact value (no decimals) of sin 5 cos 40  cos 5 sin 40 without using
your calculator except to check your answer.
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The sum of angles formula gives the result for sin(5  40 )  sin 45 
2
.
2