Download Math 1113 Practice Test 1 Fall 2010 0. (2 points if it is printed neatly

Math 1113 Practice Test 1 Fall 2010
0. (2 points if it is printed neatly) Name:______________________________
1. Sketch 210° in standard position.
210° is 30° more than 180° so, bearing in mind that one hour on a clock is 30° we have the following:
2. (3 points) Find two angles, one positive and one negative, that are coterminal with 135°
To find coterminal angles you add or subtract multiples of 360°. There are many possible answers. One
possibility is:
135° + 360°, 135° − 360°
= 495°, −215°
3. (2 points) Convert 
You multiply by
3
to degrees without using a calculator
5
180


3 180

 108
5 
4. (6 points) Find the exact values of the six trigonometric functions of 
13

12
Find the missing side using Pythagoras. If you recognize it is a 5-12-13 triangle you can just label the
side as 5. Call the side b.
122  b 2  132
144  b 2  169
b 2  25
b5
5. (6 points) Find the exact value of  in radians, 0   
(a) cos  
2
2
(b) sin  
3
2

2
(c) tan  
, without using a calculator
3
3
You use the following triangles, from which you can read off the values of the various trigonometric
functions. The first is a 45, 45,90 triangle and the second is a 30, 60,90 triangle. For example,
cos 45 

1
2
, so the answer to (a) is , (because you must answer in radians.)

4
2
2
2
1
2
3
1
1
6. (6 points) Find the exact values of the six trigonometric functions of  if (2, −3) is a point on the
terminal side of  in standard position.
You use the following picture. You do not really need to draw the picture, but it lets you see what to do.
r is the standard letter for the distance from the origin to a point.
1
r
(2, 3)
By Pythagoras, r 
sin  
x 2  y 2  22  (3) 2  13 . Then
y
3
x
2
y
3

,cos   
, tan     , and so on.
r
r
x
2
13
13
7. (6 points) Without using a calculator evaluate: (Draw the angle in standard position and use a
definition. For example, sin  
y
)
r
(a) cos 
(b) tan 270
(c) sec2
If you look on the unit circle, for which r = 1, you see (1, 0) is on the terminal side of  , (0, 1) is on the
terminal side of270 and (1, 0) is on the terminal side of 2 .
(a) cos  
x 1

 1
r 1
(b) tan 270 
y
is undefined
x
(c) sec 2 
r 1
 1
x 1
(b) tan 270 
y
is undefined
x
(c) sec 2 
1 1
 1
x 1
Note that since r = 1 you could say
(a) cos   x  1
8. (3 points) Find the reference angle for each of the following angles:
(a)   25
(b)  
7
6
(c)   480
The reference angle is the (positive) acuter angle and the terminal side of  .
9. (3 points) Put the correct sign in the box. That is, put + or – in the box.
(a) tan    tan  ',
 is in QIII
(b) sin    sin  ',
 is in QIV
(c) cos    cos  ',
 is in QIV
1
2
10. (4 points) Find two solutions of the equation cos   . Give your answers in degrees with
0    360
Since we know cos 60⁰ =
1
the reference angle is 60⁰. The angles must be in the second and third
2
quadrants because that is where the cosine is negative. The picture is:
60⁰
60⁰
  180  60 ,180  60  120 , 240
11. (4 points)A tree casts a shadow 20 feet long when the angle of elevation of the sun is 58°. How high
is the tree?
h
58⁰
20 ft
h
 tan 58
20
h  20 tan 58
h  32.007
The tree is 32 feet tall
12. (4 points) Maggie is flying her kite using 50 feet of string, holding the string next to the ground.
Sabra is 30 feet way from Maggie and notices that the kite is directly over her head. How high is the
kite?
40 feet
13. Without using a calculator graph one cycle of y  2 cos(3 x   ) . To get credit you must make it
clear how you determine the endpoints of the cycle you graph.
Period =
2 2

B
3
1
1 2 

(period)  
4
4 3
6
End points: 3x    0 giving x 

3
The graph is in a separtate document.
and 3x    2 giving x  
14. Without using a calculator graph one cycle of y  3sin(2 x   ) . To get credit you must make it
clear how you determine the endpoints of the cycle you graph.
Period =
2 2

1
B 2
1
1
1
(period)  1 
4
4
4
End points: 2 x    0 giving x  
1
2
and 2 x    2 giving x 
1
2
The graph is in a separtate document.
15. Without using a calculator graph one cycle of y  3cot(2 x 

2
) . To get credit you must make it
clear how you determine the endpoints of the cycle you graph.
Period =

B


2
1
1  
(period)   
4
4 2 8
End points: 2 x 

2
 0 giving x 

4
and 2 x 

2
  giving x 
3
4
The graph is in a separtate document.
16. Find the length of the third side of the triangle in terms of x. Then find  in terms of x for all three
inverse trigonometric functions.
5
x

  sin 1
x
5
  cos 1
25  x 2
5
  tan 1
x
25  x 2
17. Find the exact value of sin 1 (sin
Draw a picture of
7
).
4
7
7
7


) then sin   sin
: If   sin 1 (sin
, but you must have    
4
4
4
2
2
From the picture you can see that


4
is coterminal with
7
and is in
4
the right range.
Answer: 

4
18. Find the exact value of tan(arcsin( 34 ))
Let  =arcsin( 
sin  
3
) and draw a picture of  .
4
You must have sin   
3


and     . Since
4
2
2
y
make r = 4 and y =−3
r
For the point shown on the terminal
side of  you have y = −3 and r =4.
Find x using Pythagoras: x 2  y 2  r 2
x 2  (3) 2  42
x2  7
x 7
Note you can see x is positive from
the location of 
tan(arcsin( 34 )) =
y
3

x
7
x
4
18. Write as an algebraic expression tan(sin 1 )
Let   sin 1
x
x
then sin  
4
4
Draw a right triangle containing an angle who's sine is
4
x
4
x
a
By Pythagoras a 2  x 2  42
a 2  16  x 2
a  16  x 2
x
x
Then tan  = 
a
16  x 2
Questions 19 and 20 use the following figure
B
c
a
C
b
A
19. Solve the triangle if A  28 and a  12.3 . Round answers to three decimal places.
B  62 ;
b  23.133 ;
b  26.200
20. Solve the triangle if a  4 and c  7 . Round answers to two decimal places.
b  33  5.74 ;
A  34.85;
B  55.15
221. Finding the height of a Flagpole. A flagpole is mounted on the edge of a tall building. From a point
30 feet from the bottom of the building the angle of elevation of the bottom of the flagpole is 60, and
the angle of elevation of the top of the flagpole is 70. How tall is the flagpole?
30.46 ft