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Math 175: Cloutier
Exam 5
Name: _____________________________________
SP2013
ID: __________________________
Instructions:
Spread out!
Cover your tests!
Keep your eyes on your own papers!
You may use your calculators, but NO CELL PHONES!
Remove all ear pieces. Put away all electronic devices, the time will be written on the board.
Show your work on EVERY problem. Points will not be awarded without sufficient work.
You will be given FIFTY-FIVE MINUTES for this Exam.
The following formulas may be needed:
Page #
2
3
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Total
22
38
40
100
Points Earned
Possible
1
Math 175: Cloutier
Exam 5
SP2013
1. (10pts) The angle of inclination from the base of skyscraper A to the top of skyscraper B is
10.8 . If skyscraper B is 1414 feet tall, how far apart are the two skyscrapers?
1414
ft
10.8*
tan 10.80  
opp
adj
1414
tan 10.80  
x
1414
x
 7, 412.45 ft
tan 10.80 
2. (12pts) Find the value of the angle  .
Left : sin  
2
4
& Right : sin  
h2
h8
2
4

h 2 h8
Since both have sin  , put them together!
Obviously, solve for h:
2h  16  4h  8  8  2h  h  4
Now solve for theta: Use the Left , doubled : sin
2

2

2
1
   2  sin 1    38.9o
6
 3
Math 175: Cloutier
Exam 5
SP2013
3. (18pts) An aircraft is spotted in the sky between two observers who are 1900 feet apart on
flat land. The first person measures the angle of elevation to the airplane to be 35 and the
second person measures the angle of elevation to the airplane to be 30 . How high is the
airplane?
b
30*
35*
1900 ft
There’s 180* in a triangle: 180 – 35 – 30 = 115*
sin 115o 
1900

sin  30o 
b
 b
1900sin  30o 
sin 115o 
 1048.2 ft
Now use right triangle trigonometry:
1900sin  30o  sin  35o 
h
o
o
sin  35  
 h  b sin  35  
 34, 449.26 ft
b
sin 115o 
4. (20pts) Solve the triangle: b  4 , c  5 , B  40
sin  40o 
4
5sin  40
sin C

 sin C 
5
4
o
 remember that sine is positive in Q1 and Q2:
 5sin  40o  
  53.5o
Q1 answer: C  sin 


4


 5sin  40o  
  126.5o
Q2 answer: 180  sin 


4


Thus: A  180  40  53.5  86.5o
Thus: A  180  40  126.5  13.5o
1
sin  86.50 
a

sin  400 
4
 a
4sin 86.50 
sin  400 
1
 6.2
3
sin 13.50 
a

sin  400 
4
 a
4sin 13.50 
sin  400 
 1.5
Math 175: Cloutier
Exam 5
SP2013
5. A ship travels 12 miles per hour. Its original destination was 600 miles away. To
avoid a storm, the ship traveled 7 hours off course, at an angle of 31 .
a. (20pts) Solve the shown triangle:
12miles / hr  v
v
s
s
 12 
 s  84miles
t
7
b2  6002  842  2  600 84  cos  310   b  529.8miles
sin  31 sin  C 

 C  4.68o
529.8
84
And hence A  180  4.68  31  144.32o
b. (5pts) Find the area of the triangle you solved above.
1
1
A  ab sin C   600  529.8 sin  4.68o   12,968mi 2
2
2
6. (15pts) Find the area of the following triangle: a  7 , b  5 , c  8
82  7 2  52  2  7  5  cos  C 
64  49  25
70
1
1
64  49  25
Area  ab cos C   7  5 
 2.5
2
2
70
cos  C  
cos 
sin 

 sin   cos 
1  tan  1  cot 
cos 
sin 
cos 
sin 
cos 2 
sin 2 
cos2 
sin 2 







1  tan  1  cot  1  sin  1  cos  cos   sin  sin   cos  cos   sin  cos   sin 
cos 
sin 
2
2
cos   sin   cos   sin   cos   sin  


 cos   sin 
cos   sin 
cos   sin 
Bonus: Establish the identity:
4