Download Pre-Class Problems 14 for Tuesday, November 4, and Thursday

Pre-Class Problems 14 for Tuesday, November 4, and Thursday, November 6
These are the type of problems that you will be working on in class. These
problems are from Lesson 9.
Solution to Problems on the Pre-Exam.
You can go to the solution for each problem by clicking on the problem letter.
Objective of the following problems: To find the value of the inverse cosine of a
given number.
1.
Find the exact value of the following.
c.

2 


Arc cos 

2 

cos  1 1
f.
Arc cos 0
Arc cos (  1 )
i.
 1
cos  1   
 2
a.
1
Arc cos
2
b.
cos
d.

3

Arc cos  
 2 


e.
g.
cos  1
2
2
h.
1
3
2
Objective of the following problems: To use an inverse trigonometric function to
find the exact value of an angle in a right triangle and to use technology to
approximate the angle.
2.
Find the exact value of the following angles. Then approximate the angle to
the nearest tenth of a degree.
a.
b.

32
7
45

12
c.
A 15-foot ladder is leaning against the top of a vertical wall. If the
bottom of the ladder is 10 feet from the base of the wall, then find the
angle of depression from the top of the ladder to the ground.
d.
An observer is 70 meters from the base of a 150-meter building. What
is the angle of elevation from the observer to the top of the building?
Objective of the following problems: To use the inverse tangent function to find
the reference angle of an angle if given the coordinates of a point on the terminal
side of the angle.
3.
Find the exact and the approximate angle  , which is between 0  and 360  ,
which passes through the following points. Round your approximate answer
to the nearest tenth of a degree.
a.
(  3, 7 )
b.
(10 ,  6 )
d.
( 5 , 2)
e.
(12 , 
c.
(  5,  8)
3)
Additional problems available in the textbook: Page 186 … 7, 8, 11, 15, 103, 104,
107 - 109. Example 3a on page 183. Page 196 … 9 - 12, 27 - 30. Example 4 on
page 191.
Solutions:
1
Definition The inverse cosine function, denoted by cos , is defined by
y  cos  1 x if and only if cos y  x , where  1  x  1 and 0  y   .
1
Notation: Sometimes, cos x is denoted by Arc cos x .
cos  1 x .
That is, Arc cos x =
The restriction that is put on the cosine function in order to make it be one-to-one
means that your angle answer must be in the closed interval [ 0 ,  ] . NOTE: The
 
angles in the open interval  0 ,  are in the first quadrant. The angles in the open
 2


interval  ,   are in the second quadrant. The angle 0 is on the positive x-axis,
2 
the angle
1a.

is on the positive y-axis, and the angle  is on the negative x-axis.
2
Arc cos
1
2
Answer:

3
1
NOTE: Since
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open

1
 

interval  0 ,  .
We know that cos
either by Unit Circle
3
2
 2
 1
Trigonometry or Right Triangle Trigonometry. Note that the point  ,  is
 3 2
on the graph of the cosine function.
Back to Problem 1.
1b.
cos  1
3
2
Answer:

6
3
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open
3

 

interval  0 ,  . We know that cos
either by Unit Circle
6
2
 2
Trigonometry or Right Triangle Trigonometry.
Note that the point

3
 ,

 6 2  is on the graph of the cosine function.


NOTE: Since
Back to Problem 1.
1c.

2 

Arc cos  


2


Answer:
3
4
2
is negative and not minimum negative of  1 , then the
2
angle answer is in the second quadrant. That is, the angle answer is in the


open interval  ,   .
2 
NOTE: Since 

2 
2





Arc
cos


'

Arc
cos

If we let
,
then
. Thus,

2 
2
4


2 
4

3

  3 .
Arc
cos

    ' 


. That is,

2 
4
4
4
4

2

Arc
cos

NOTE: We will show that
in Problem 1g.
2
4
2
3

  cos
 
using reference angles. Note
4
4
2
 3
2 

 is on the graph of the cosine function.
,

that the point  4
2 

We know that cos
Back to Problem 1.
1d.

3

Arc cos  

2 

Answer:
5
6
3
is negative and not minimum negative of  1 , then the
2
angle answer is in the second quadrant. That is, the angle answer is in the
NOTE: Since 


,


.
open interval
2 

3
3





Arc
cos


'

Arc
cos

If we let
,
then
. Thus,

2 
2
6


3
5
6

5


Arc
cos


    ' 


. That is,

2 
6 .
6
6
6

3

Arc
cos

NOTE: We showed that
in Problem 1b.
2
6
3
5

  cos
 
using reference angles. Note
6
6
2
 5
3

 is on the graph of the cosine function.
,

that the point  6
2 

We know that cos
Back to Problem 1.
1e.
cos  1 1
Answer: 0
NOTE: Since 1 is the maximum positive of 1, then the angle answer is not in
the first quadrant. So, the angle answer is on one of the coordinate axes,
namely the positive x-axis. The angle which is in the closed interval [ 0 ,  ]
and lies on the positive x-axis is the angle 0. We know that cos 0  1 by
Unit Circle Trigonometry. Note that the point ( 0 , 1 ) is on the graph of the
cosine function.
Back to Problem 1.
1f.
Arc cos 0

Answer:
2
NOTE: Since 0 is not positive, then the angle answer is not in the first
quadrant. Since 0 is not negative, then the angle answer is not in the second
quadrant. So, the angle answer is on one of the coordinate axes, namely the
positive y-axis. The angle which is in the closed interval [ 0 ,  ] and lies on


 0 by Unit Circle
. We know that cos
2
2
 
Trigonometry. Note that the point  , 0  is on the graph of the cosine
2 
function.
the positive y-axis is the angle
Back to Problem 1.
1g.
cos  1
2
2
Answer:

4
2
is positive and not maximum positive of 1, then the angle
2
answer is in the first quadrant. That is, the angle answer is in the open
NOTE: Since
2

 
0
,
cos



interval
. We know that
either by Unit Circle
4
2
 2
Trigonometry or Right Triangle Trigonometry. Note that the point

2 
 ,

 4 2  is on the graph of the cosine function.


Back to Problem 1.
1h.
Arc cos (  1 )
Answer: 
NOTE: Since  1 is the minimum negative of  1 , then the angle answer is
not in the second quadrant. So, the angle answer is on one of the coordinate
axes, namely the negative x-axis. The angle which is in the closed interval
[ 0 ,  ] and lies on the negative x-axis is the angle  . We know that
cos    1 by Unit Circle Trigonometry. Note that the point (  ,  1 ) is on
the graph of the cosine function.
Back to Problem 1.
1i.
 1
cos  1   
 2
Answer:
2
3
1
is negative and not minimum negative of  1 , then the
2
angle answer is in the second quadrant. That is, the angle answer is in the


open interval  ,   .
2 
NOTE: Since 
1 
If we let   cos  

3
    ' 

3
1

1 1
 . Thus,
 , then  '  cos
2
2
3
1
2

2
1 

. That is, cos    
.
3
3
3
 2
1
NOTE: We showed that cos
1


in Problem 1a.
2
3
2

1
  cos
  using reference angles. Note that
3
3
2
1
 2
,   is on the graph of the cosine function.
the point 
2
 3
We know that cos
2a.
Back to Problem 2.

32
12
tan  
12
3
3

   tan  1 since  is an acute angle
32
8
8
NOTE: tan  
Answer:
opp
adj
1
Exact:   tan
3
8
Approximate:   20.6 
2b.
Back to Problem 2.
45
7

sin  
45
45
   sin  1
since  is an acute angle
7
7
NOTE: sin  
Answer:
opp
hyp
1
Exact:   sin
45
7
Approximate:   73.4 
2c.
A 15-foot ladder is leaning against the top of a vertical wall. If the bottom of
the ladder is 10 feet from the base of the wall, then find the angle of
depression from the top of the ladder to the ground.
Let  be the angle of depression.
---------------

15 feet

x
10 feet
10
2
2

   cos  1 since  is an acute angle
15
3
3
cos  
NOTE: cos  
Answer:
adj
hyp
1
Exact:   cos
2
3
Approximate:   48.2 
2d.
Back to Problem 2.
An observer is 70 meters from the base of a 150-meter building. What is the
angle of elevation from the observer to the top of the building?
Let  be the angle of elevation from the observer to the top of the building
Top of Building
150 meters

Observer
70 meters
tan  
150 15
15

   tan  1
since  is an acute angle
70
7
7
NOTE: tan  
opp
adj
1
Exact:   tan
Answer:
15
7
Approximate:   65.0 
3a.
Back to Problem 2.
(  3, 7 )
tan  
Back to Problem 3.
y
7
 
x
3
NOTE: Since the point (  3 , 7 ) is in the second quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the second
quadrant. Thus,  is not an acute angle.
To find the reference angle  ' : tan   
 '  tan  1
7
since  ' is an acute angle.
3
 '  tan  1
7
 66.8 
3
7
7
 tan  ' 

3
3
NOTE: Since  is in the second quadrant, then   180    ' . Since the
1 7
1 7
exact value of  ' is tan
, then the exact value of  is 180   tan
.
3
3
Since the approximate value of  ' is 66.8  , then the approximate value of 
is 180   66.8   113.2  .
Answer:
1
Exact:   180   tan
Approximate:   113.2 
3b.
(10 ,  6 )
7
3
Back to Problem 3.
tan  
y
6
3
 
 
x
10
5
NOTE: Since the point (10 ,  6 ) is in the fourth quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the fourth
quadrant. Thus,  is not an acute angle.
To find the reference angle  ' : tan   
 '  tan  1
3
since  ' is an acute angle.
5
 '  tan  1
3
 31.0 
5
3
3
 tan  ' 

5
5
NOTE: Since  is in the fourth quadrant, then   360    ' . Since the
1 3
1 3
exact value of  ' is tan
, then the exact value of  is 360   tan
.
5
5
Since the approximate value of  ' is 31.0  , then the approximate value of 
is 360   31.0   329.0  .
3
5
Approximate:   329.0 
1
Exact:   360   tan
Answer:
3c.
(  5,  8)
tan  
Back to Problem 3.
y
8

x
5
NOTE: Since the point (  5 ,  8 ) is in the third quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the third
quadrant. Thus,  is not an acute angle.
To find the reference angle  ' : tan  
 '  tan  1
8
since  ' is an acute angle.
5
 '  tan  1
8
 58.0 
5
8
8
 tan  ' 

5
5
NOTE: Since  is in the third quadrant, then   180    ' . Since the
1 8
1 8
exact value of  ' is tan
, then the exact value of  is 180   tan
.
5
5
Since the approximate value of  ' is 58.0  , then the approximate value of 
is 180   58.0   238.0  .
8
5
Approximate:   238.0 
1
Exact:   180   tan
Answer:
3d.
( 5 , 2)
tan  
Back to Problem 3.
y

x
2
5
NOTE: Since the point ( 5 , 2 ) is in the first quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the first
quadrant.
NOTE: By the definition of the inverse tangent function, the inverse tangent
 
of a positive number is an angle in the open interval  0 ,  .
2

tan  
2
2
   tan  1
is an acute angle.
5
5
  tan  1
2
5
Approximate:   41.8 
1
Exact:   tan
Answer:
3e.
2
 41.8 
5
(12 , 
3)
tan  
3
y
 
x
12
Back to Problem 3.
NOTE: Since the point (12 ,  3 ) is in the fourth quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the fourth
quadrant. Thus,  is not an acute angle.
To find the reference angle  ' : tan   
 '  tan  1
3
since  ' is an acute angle.
12
 '  tan  1
3
 8.2 
12
3
3
 tan  ' 

12
12
NOTE: Since  is in the fourth quadrant, then   360    ' . Since the
3
1
exact value of  ' is tan
, then the exact value of  is
12
3
360   tan  1
. Since the approximate value of  ' is 8.2  , then the
12
approximate value of  is 360   8.2   351.8  .
Answer:
1
Exact:   360   tan
Approximate:   351.8 
3
12
Solution to Problems on the Pre-Exam:
16.
Back to Page 1.
Find the exact value of each of the following. (3 pts. each)
b.

2 

cos  1  


2


3
Answer:
4
2
is negative and not minimum negative of  1 ,
2
then the angle answer is in the second quadrant. That is, the angle


,


.
answer is in the open interval
2 
NOTE: Since 

2 
2

1 
1



cos


'

cos

If we let
,
then
. Thus,

2 
2
4


2  3
4

3
1 
 
cos

    ' 


. That is,


2
4 .
4
4
4


2
3

  cos
 
using reference angles.
4
4
2
 3
2 

 is on the graph of the cosine function.
,

Note that the point  4

2


We know that cos
e.
Arc cos
1
2
Answer:

3
1
is positive and not maximum positive of 1, then the
2
angle answer is in the first quadrant. That is, the angle answer is in the

1
 

open interval  0 ,  . We know that cos
either by Unit Circle
3
2
 2
Trigonometry or Right Triangle Trigonometry. Note that the point
 1
 ,  is on the graph of the cosine function.
 3 2
NOTE: Since
14.
Find the exact angle  , where 0     360  , if the terminal side of 
passes through the point (  6 ,  14 ) . (5 pts.)
tan  
y
14
7


x
6
3
NOTE: Since the point (  6 ,  14 ) is in the third quadrant and lies on the
terminal side of the angle  , then the terminal side of  is in the third
quadrant. Thus,  is not an acute angle.
To find the reference angle  ' : tan  
 '  tan  1
7
7
 tan  ' 

3
3
7
since  ' is an acute angle.
3
NOTE: Since  is in the third quadrant, then   180    ' . Since the
1 7
1 7
exact value of  ' is tan
, then the exact value of  is 180   tan
.
3
3
Answer:
  180   tan  1
7
3