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Right triangle trigonometry definitions.
Build a right triangle in the diagram over
the central angle.
Label the right triangle.
c is the
Label the right triangle using
theta(  ) as the reference angle.
This angle is always ACUTE!
B
c
a
HYPOTENUSE.
OPPOSITE of  .

A
The length Opposite of  .

The length of the Hypotenuse.
a is the side
b is thebside
C
ADJACENT to 
.
a
sin  

c
The length Adjacent of  .
b
cos  

The length of the Hypotenuse.
c
The length Opposite of  .
a
tan   

The length Adjacent of  .
b
SOH-CAH-TOA
I P Y
N P P
E O O
S T
I E
T N
E U
S
E
O
S
I
N
E
D
J
A
C
E
N
T
Y
P
O
T
E
N
U
S
E
A
N
G
E
N
T
P
P
O
S
I
T
E
D
J
A
C
E
N
T
Right triangle trigonometry definitions.
Build a right triangle in the diagram over
the central angle.
Label the right triangle.
c is the
Label the right triangle using
theta(  ) as the reference angle.
This angle is always ACUTE!
a2  b2  c2
B
c
HYPOTENUSE.
Pythagorean
Theorem
a a is the side
ADJACENT
OPPOSITE of  .

A
b
b is the side
The length Opposite of  .

The length of the Hypotenuse.
REMEMBER
C
ADJACENT of  .
OPPOSITE
ba
sin  

c
The length Adjacent of  .
ba
cos  

The length of the Hypotenuse.
c
The length Opposite of  .
ba
tan   

The length Adjacent of  .
ba
SOH-CAH-TOA
I P Y
N P P
E O O
S T
I E
T N
E U
S
E
O
S
I
N
E
D
J
A
C
E
N
T
Y
P
O
T
E
N
U
S
A
N
G
E
N
T
P
P
O
S
I
T
E
D
J
A
C
E
N
T
Find the 6 trigonometric functions with respect to  .
Circle the reference angle and label the
opposite, adjacent, and hypotenuse.
Find the value of the missing side, c.
a2  b2  c2
3 4 c
2
2
2
HYP.
c 5
c 5
OPP.
9  16  c 2
25  c 2
SOH-CAH-TOA
3
sin   
5
4
cos  
5
5
csc  
3
5
sec   
4
3
tan   
4
4
cot   
3
ADJ.
b
Special Right Triangle Relationships.
b2  b2  c2
2b  c
2
2
c  2b
2
45
2
cb 2
b
b
45 : 45 : 90
b:b:b 2
45
b
1:1: 2
1
2
sin 45 

2
2
2
cos45 
2
tan 45  1
c b 2
45
csc45  2
2
1
45
sec45  2
cot45  1
1
These answers are
considered exact values.
Special Right Triangle Relationships.
a 2  b 2  2a 
2
a 2  b 2  4a 2
b  3a
2
60
30
b 2  3a 2
ba 3
2a
2a
ba 3
2
30 : 60 : 90
a : a 3 : 2a
1: 3 : 2
60
a
a
2a
60
1
sin 30 
2
csc30  2
30
3
cos30 
2
2
2 3
sec30 

3
3
1
3
tan 30 

3
3
cot 30  3
2
3
60
1
3
sin 60 
2
1
cos60 
2
tan 60  3
csc 60 
2 3
3
sec60  2
cot 60 
3
3
Do you see a pattern?
Cofunction Identities.
Two positive angles are complimentary if
their sum is 90o. Our trigonometric
functions are identified with the prefix “Co”.
Sine & Cosine
Tangent & Cotangent
Secant & Cosecant
sin    cos90     ac
cos   sin 90     bc
tan    cot90     ba
cot   tan 90     ba
sec   csc90     bc
csc   sec90     ac
From the 30-60-90 Ex.
1
sin 30 
2
1
cos90  30  cos60 
2
cos   sin 90   
cos52  sin 90  52
cos52  sin 38
tan    cot 90   
tan 71  cot90  71
tan 71  cot19
sec   csc90   
sec24  csc90  24
sec24  csc66
sin    cos90   
sin   15  cos90    15
sin   15  cos75   
csc90    10  csc2  20
90    10  2  20
80    2  20
80  3  20
60  3
20  
Definition of Reference Angle.
Let  be a nonacute angle in standard position that lies in a quadrant. Its
reference angle is the positive acute angle   formed by the terminal side of
and the x – axis.

Find all six trigonometric function exact values of 210.
•Use special right triangles and reference angles to find the exact
values of the trigonometric functions. 30 : 60 : 90 45 : 45 : 90
1: 3 : 2
1:1: 2
opp
1
SOH – CAH - TOA
sin 210 

hyp
cos210 
2
adj
 3

hyp
2
tan 210 
1
3
opp


adj
3
 3
cot 210 
90
S
A
adj
210
180  3
30
1
opp
2
T
Reference angle
hyp
 3
adj

 3
1
opp
hyp
2
2 3
sec210 


adj
3
 3
csc210 
hyp
2

 2
opp  1
C
Use special right triangles and reference angles to find the exact
values of the trigonometric functions. 30 : 60 : 90 45 : 45 : 90
1: 3 : 2
1:1: 2
SOH – CAH - TOA
cos 240
S
opp
hyp
2
3
60
1
T
A
adj
Reference angle
tan 675
S
675
 240
C
1
adj

cos 240 
hyp
2
T
tan  675  
A
adj1
Reference angle
45
opp
2
hyp
1
C
opp
1

 1
adj
1
S
opp
A
hyp
2
3
2
adj
C
  225
 135
45
45
C
S
A
adj
3
30
60
1
Reference angle
A
T
opp
3
T
adj
 3
1

 2

2
2


2
1  3   1 
 2     
3
2
 4 3
S
A
hyp
120
60
1
T
S
C
2
 135
  225
hyp
T
 1 


3


2
opp
1
C
2
Give away
for 45o angle
MODE
Right now we want DEGREE mode, move cursor to DEGREE and hit ENTER
2nd APPS activate the ANGLE window.
x-1 button, Sine, Cosine, and Tangent
Degree symbol.
Minute symbol.
are to the right.

0.7571217563
1
cos97.977
-7.205879213
 tan 51.4283
-0.4067366431
  sin 1 0.96770915
  75.4  7524'
1.253948151
1
1.0545829

cos 
1
1
Second symbol.
cos  
 18.51470432
 1830'53"
1.0545829
ALPHA, +
F  W sin    2500 sin 2.5  109 lb
F  W sin    5000sin  6.1  531 lb
right
acute
C
A
B
complimentary
B
hyp.
Find angles first. 90  34.5  55.5
SOH-CAH-TOA
a
sin 34.5 
12.7
a  12.7  sin 34.5
b
cos34.5 
12.7
b  12.7  cos34.5
c  12.7
A
34.5
b
adj.
B  ____
55.5
7.2 in
a  _____
10.5 in
b  _____
opp.
a
C
Need to find the following:
hyp.
opp.
= 29.43
= 53.58
adj.
A  _____
33.3
90  33.3  56.7
56
.
7

B  _____
44.77
b  _____
A general rule is to always use the information you are given. We can find b by
the Pythagorean Theorem.
a b  c
2
2
2
b  c2  a2
b  53.582  29.432
Sides were given to 2 decimal
places … so b = 44.77
SOH-CAH-TOA
29.43
sin  A 
53.58
 29.43 
A  sin 1 

 53.58 
Make sure the MODE is
DEGREE.
SOH-CAH-TOA
hyp.
25
tan 20 
x
25
x  tan 20   x
x
tan 20
25
x

tan 20 tan 20
x
25
tan 20
y =opp.
57.635
40 x =adj.
h
68.687
20 adj.
25 opp.
hyp.
y
tan 40 
68.687
68.687  tan 40  y
h  25  57.635
h  82.635 ft
When a single angle is given, it is understood that the
Bearing is measured in a clockwise direction from due north.
N
N
N
45o
165o
225o
Starts at a Bearing starting on the north-south line
and uses an acute angle to show the direction, either east or west.
S 45o E
N 75o W
N
45o
S
75o
C
adj. x
61o
29 opp.
29
61
29
A
61
3.7mi
SOH-CAH-TOA
331o
B
hyp.
x
cos29 
3.7
x
3.7  cos29 
 3.7
3.7
x  AC  3.2mi
C
47 43o
3.5(22) = 77
4(22) = 88
47o
P
43
c
s2  p2  c2
77 2  882  c 2
S
c  772  882  117 nautical miles
h  27
hyp.
hyp.
adj.
SOH-CAH-TOA
h
tan 15 
x  50
x  50 tan 15 
h
 x  50
x  50
h  x  50 tan 15
opp.
adj.
Solve for x to find h.
h
tan 28 
x
h  x  tan 28
h
x  tan 28   x
x
50  tan 15
x
tan 28  tan 15
x  50 tan15  x  tan28
x  tan 15  50  tan 15  x  tan 28
 x  tan 15
 x  tan 15
50  tan 15  x  tan 28  x  tan 15
50  tan 15  xtan 28  tan 15
tan 28  tan 15 tan 28  tan 15
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