Download 4.3 Right Triangle Trigonometry

4.3 Right Triangle Trigonometry
The six trigonometric functions.
Our second look at the trigonometric functions is from a right triangle perspective. Consider a right
triangle, with one acute angle labeled  , the three sides of the triangle are the hypotenuse, the
opposite side (the side opposite the angle  , and the adjacent side (the side adjacent to the angle  ).
Right Triangle Definitions of Trigonometric Functions.
opposite
hypotenuse
hypotenuse
csc  
opposite
sin  
adjacent
hypotenuse
hypotenuse
se c  
adjacent
cos  
opposite
adjacent
adjacent
coc 
opposite
tan  
Since any two right triangles with angle  are similar, the trigonometric ratios are the same, regardless
of the size of the triangle; the trigonometric ratios depend only on the angle  .
Example: Find the six trigonometric functions of angle  for the following graph.
We have
7
3
,cos   , tan  
4
4
4
4
csc  
,sec   ,cot  
3
7
sin  
7
3
3
7
Example: Consider a right triangle with  as one of its acute angles. If tan  
sin  ?
3
, find the value of
2
Solution: Construct a right triangle having a as one of its acute angles. Since tan  
3
, we can choose
2
the opposite and adjacent sides to be 3 and 2.
So, the hypotenuse= 32  22  13
Thus, sin  
3
13
Question: Consider a right triangle with  as one of its acute angles. If sin  
 5
tan  ?  
 12 
Special Triangles.
5
, find the value of
13
 in degrees
30
sin 
1
2
 in radians

6

4

3
45
60
2
2
3
2
cos 
tan 
3
2
2
2
1
2
3
3
1
3
In the box, note that sin 30  cos60 . This occurs because 30 and 60 are complementary angles. In
general, it can be shown fromt eh right triangle definitions that cofunctions of complementary angles
are equal. That is, if  is an acute angle, the following relationships are true.
sin(90   )  cos  ,cos(90   )  sin  , tan(90   )  cot 
sec(90   )  csc  ,csc(90   )  sec  ,cot(90   )  tan 
Trigonometric Identities.
In trigonometry, a great deal of time is spent studying relationships between trigonometric functions
(identities).
Fundamental Trigonometric Identities.
sin  
1
,
csc 
tan  
sin 
,
cos 
cos  
1
,
se c 
cot  
sin 2   cos2   1,
tan  
1
cot 
cos 
sin 
1  tan 2   sec 2  ,
1  cot 2   cs c 2 
Example: Let  be an acute angle such that sin  
5
. Find the values of cos and tan  .
13
cos2   sin 2   1
2
2
25
144
12
5
5
cos    1  sin 2    1      1      1 


169
169
13
 13 
 13 
Since  is an acute angle, so cos  
12
13
5
sin  13 5 13 5
tan  

  
cos  12 13 12 12
13
Example. Let  be an acute angle such that tan   2 . Find the value of sin  .
tan   2
1
cot  
2
2
1  cot   cs c 2 
2
1 5
1
cs c 2   1     1  
4 4
2
1
5

2
sin  4
4
sin 2  
5
4
sin   
5
Since  is an acute angle, so sin  
4
5
Applications of Trigonometry of Right Triangles.
Example: Solve the following triangle ABC .
Solution:
sin 30 
a
12
a  12sin 30  12 
cos 30 
1
6
2
b
12
b  12 cos 30  12 
3
6 3
2
Example: A giant tree casts a shadow 500 ft long. Find the height of the tree if the angle of elevation of
the sun is 25.7 .
Solution: Let the height of the tree be h .
h
500
h  500  tan 25.7  500  0.48127  240.6
tan 25.7 
The height of the tree is about 240.6 ft.
Example: From a point on the ground 500 ft from the base of building, and observer finds that the angle
of elevation to the top of the building is 24 and the angle of elevation to the top of flagpole atop the
building is 27 . Find the height of the building and the length of the flagpole.
Solution:
h
500
h  500  tan 24  500  0.4452  223
k
tan 27 
500
k  500  tan 27  500  0.5095  255
k  h  255  223  32
tan 24 
Length of the pole is approximately 32 ft.