Download 1.1 Trigonometric Functions of Acute Angles Solutions

1.1 Trigonometric Functions of Acute Angles Solutions
Find the six trigonometric function values of the specified angle
1.
2.
3 3

15
17
3
6

8
sin  
15
17
csc  
17
15
sin  
3 1

6 2
csc  
cos  
8
17
sec  
17
8
cos  
3 3
3

6
2
sec  
2
2 3

3
3
tan  
15
8
cot  
8
15
tan  
cot  
3

1
3
3 3

1
3

3
3
2
 2
1
3
Given a function value of an acute angle, find the other five trigonometric function values.
3. sin  
24
25
4. tan   2
25
2

24
1
Again use Pythagorean Theorem to

Find the missing side using the
Pythagorean Theorem and obtain 7
find the missing side and obtain
sin  
24
25
csc  
25
24
sin  
2
2 5

5
5
cos  
7
25
sec  
25
7
cos  
1
5

5
5
tan  
24
7
cot  
7
24
tan   2
Find the exact function value (do not use a calculator)
5. cos 450
6. sec600
7. cot 600
8. sin 300
5
2
csc  
sec  
5

1
cot  
1
2
9. tan 450
5
5
10. csc300
2
3
1
2
1
2
2
2
3
Note: Just use the two special triangles that you are supposed to know along with your trigonometric
function definitions to figure these out!
Convert to decimal notation. Round to 2 decimal places.
11. 90 43
12. 4903846
0
0
 38   46 
0
49 3846  490     
  49.65
 60   3600 
0
 43 
9 43  9     9.720
 60 
0
0
0
Convert to degrees, minutes and seconds. Round to the nearest second.
13. 17.6420
 60 
 60 
0
0
17.6420  170  .6420  170  .6420  0   17038.52  17038  .52  17 038  .52 
  17 3831.2  17 3831

1
1




Find the function value using a calculator. Round to 4 decimal places.
14. cos510
15. tan 4013
.6293
.0737
Find the acute angle  , to the nearest tenth of a degree.
16. sin   .5125
sin   .5125    sin 1 .5125  30.80
Find the exact acute angle  . Do not use a calculator.
17. cos  
3
2
3
2
 3
  cos 1 

 2 
we are looking for THE acute angle
(that we are supposed to know)
cos  
3
.
2
If you know your special triangles
whose cosine is
then you know the answer is 300
18. csc  
csc  
2 3
3
2 3
3
 sin  
3
2 3

3
2 3

3 3 3
3


6
2
3
so we are looking for the angle whose sine is
Again, if you know your special triangles you
will realize that the angle we are looking for
is 600
3
.
2