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Trigonometry
Pythagoras Theorem &
Trigo Ratios of Acute
Angles
Pythagoras Theorem
a 2 + b2 = c 2
where c is the hypotenuse while a
and b are the lengths of the other two
sides.
c
a
b
Trigo Ratios of Acute angles
P
hypotenuse
opposite

O
adjacent
Q
Hypotenuse = side opposite right angle/longest side
Adjacent = side touching theta
Opposite= side opposite theta
Trigo Ratios of Acute angles
Hypotenuse = AB
A
Adjacent = AC

Opposite= BC
B
C
X

Hypotenuse = XZ
Y
Adjacent = XY
Opposite= YZ
Z
Trigo Ratios of Acute angles
P
hypotenuse
opposite

O
adjacent
Tangent ratio
tan
Q
Cosine ratio
Sine ratio
cos
sin
Trigo Ratios of Acute angles
P
hypotenuse
opposite

O
adjacent
Q
opposite PQ
tan  

adjacent OQ
adjacent
OQ
cos 

hypotenuse OP
opposite
PQ
sin  

hypotenuse OP
Trigo Ratios of Acute angles
P
hypotenuse
opposite

O
adjacent
TOA CAH SOH
Q
opposite PQ
tan  

adjacent OQ
adjacent
OQ
cos 

hypotenuse OP
opposite
PQ
sin  

hypotenuse OP
Exercise 1
13
5

sin  
opposite
5

hypotenuse 13
cos  
adjacent
12

hypotenuse 13
tan  
12
opposite 5

adjacent 12
opposite
4
sin  

hypotenuse 5
5

4
3
adjacent
3
cos  

hypotenuse 5
opposite 4
tan  

adjacent 3
Exercise 1
12
9

sin  
opposite
9 3
 
hypotenuse 15 5
cos  
adjacent
12 4
 
hypotenuse 15 5
15
tan  
opposite
24 8
sin  


hypotenuse 51 17
24

51
opposite 9 3
 
adjacent 12 4
45
adjacent
45 15
cos  


hypotenuse 51 17
opposite 24 8
tan  


adjacent 45 15
Exercise 2
3
sin A 
5
5
A
cos A 
4
5
tan A 
3
4
3
4
8
cos B 
17
17
B
sin B 
15
17
tan B 
15
8
sin C 
21
29
cos C 
20
29
15
8
21
tan C 
20
29
C
20
21
Exercise 3
sin 60  0.866

cos 60  0.5

tan 45  1

sin 0.5  0.00873

cos 30  0.866

tan 23  0.424

Exercise 4
sin (30°  2) = 0.2588…
sin (  2) = sin   2
cos 2 = 2  cos 
sin 30°  2 = 0.25
cos (2× 30°) = 0.5
2× cos 30° = 1.732…
tan (A + B) = tan A + tan B
tan (10° + 30°) = 0.839…
tan 10° + tan 30° = 0.75
Exercise 5
sin  = 0.4537
 = sin-1 0.4537 = 26.981≈27.0°
cos  = 0.3625
 = cos-1 0.3625 = 68.746≈68.7°
tan  = 4.393
 = tan-1 4.393 = 77.176≈77.2°
Exercise 5
sin  = 0.8888
 = sin-1 0.8888 = 62.722≈62.7°
cos  = 0.9999
 = cos-1 0.9999 = 0.8102≈0.8°
tan  = 0.5177
 = tan-1 0.5177 = 27.370≈27.4°
Further Examples 1
In the diagram, BCE is a straight line,
angle ECD = 54.8° and angle CDE =
angle ACB = 90°.
BC = 7 cm and AC = CE = 8 cm.
Calculate
B
7 cm
A
C
8 cm
angle CED,
angle DCB,
angle BAC,
the length of ED,
the length of AE,
54.8°
D
8 cm
E
Further Examples 1
In the diagram, BCE is a straight line,
angle ECD = 54.8° and angle CDE =
angle ACB = 90°.
BC = 7 cm and AC = CE = 8 cm.
Calculate
B
7 cm
A
C
8 cm
angle CED?
54.8°
D
8 cm
angle CED = 180° − 90° − 54.8° = 35.2°
E
Further Examples 1
In the diagram, BCE is a straight line,
angle ECD = 54.8° and angle CDE =
angle ACB = 90°.
BC = 7 cm and AC = CE = 8 cm.
Calculate
B
7 cm
A
C
8 cm
angle DCB?
54.8°
D
8 cm
angle DCB = 180° − 54.8° = 125.2°
E
Further Examples 1
angle BAC?
B
Let angle BAC be .
opposite
7
sin  

hypotenuse ?
7 cm
A

C
8 cm
adjacent
8
cos  

hypotenuse ?
54.8°
D
8 cm
opposite 7
tan  

adjacent 8
E
7
  tan
 41.185...  41.2
8
1
Further Examples 1
the length of ED?
B
opposite
ED
sin 54.8 

hypotenuse
8

adjacent
CD

cos 54.8 

hypotenuse
8
7 cm
A
C
8 cm
54.8°
D
8 cm
opposite ED
tan 54.8 

adjacent CD

ED
sin 54.8 
8
ED  8  sin 54.8  6.537...  6.54cm

E
Further Examples 1
the length of AE?
B
opposite
sin? 
hypotenuse
adjacent
cos? 
hypotenuse
7 cm
A
C
8 cm
54.8°
D
8 cm
opposite
tan? 
adjacent
AE  8  8
2
2
2
AE  128  11.3137...  11.3cm
E
Further Examples 2
John is flying a kite whose string is making a
angle with the ground.
The kite string is 65 meters long. How far is the kite above the ground?
Let the height be h.
sin 70 
h
65
65 m
?
cos 70 
65

tan 70 
h
?
GROUND
h
Further Examples 2
John is flying a kite whose string is making a
angle with the ground.
The kite string is 65 meters long. How far is the kite above the ground?
Let the height be h.
h
sin 70 
65

h  65  sin 70

 61.0800...
 61.1m
65 m
GROUND
?
Further Examples 3
1 foot = 12 inches
A 16 feet ladder is leaning against a house. It touches the bottom of a
window that is 12 feet 6 inches above the ground.
What is the measure of the angle that the ladder forms with the
ground?
Let the angle be .
sin  
150
192
cos  
?
192
150
tan  
?
192
16 feet
Inches

150
12 feet
inches
6 inches
Further Examples 3
1 foot = 12 inches
A 16 feet ladder is leaning against a house. It touches the bottom of a
window that is 12 feet 6 inches above the ground.
What is the measure of the angle that the ladder forms with the
ground?
Let the angle be .
sin  
150
192
 150 

  sin 

51
.
375

51
.
4

 192 

1
192 Inches

150 inches
Exercise
In the diagram, angle ADC = 30°, angle ACB = 50°,
angle ABD = 90° and BC = 4 cm. Calculate
(a) angle DAC
A
D
50°
30°
C
4 cm
B
Applications – Angle of elevation
and Angle of depression
Applications – Angle of elevation
and Angle of depression
Example 1
opposite PQ

adjacent OQ
adjacent
OQ
cos  

hypotenuse OP
opposite
PQ
sin  

hypotenuse OP
tan  
c
tan 35 
200
c  200 tan 35
 140 .04...
 140 m

Example 2
A surveyor is 100 meters from the base of a dam. The angle of elevation to the
top of the dam measures
. The surveyor's eye-level is 1.73 meters above the
ground. Find the height of the dam.
h  100 tan 26  1.73
 50.5m
At the point P, a boat observes that the angle of elevation of the cliff
at point T is 32o, and the distance PT is 150m. It sails for a certain
distance to reach point Q, and observes that the angle of elevation of
the point T becomes 48o.
T
150m
48o
32o
P
Q
R
(i) Calculate the height of the cliff.
(ii) Calculate the distance the boat is from the cliff at point Q.
(iii) Calculate the distance travelled by the boat from point P to point Q.
T
150m
48o
32o
P
Q
Let the height of the cliff = TR
TR
sin 32 
150
TR  150  sin 32  79.5m

R
T
150m
48o
32o
P
Q
R
Let the distance the boat is from the cliff at point Q = QR
TR
tan 48 
QR

TR  QR  tan 48
TR
QR 
 71.6m

tan 48
T
150m
48o
32o
P
Q
R
Let the distance travelled by the boat from point P to point Q = PQ
TR
tan 32 
PR
TR  PR  tan 32
TR
PR 

tan 32

PQ  PR  QR
 127.207  71.571
 55.6m
Trigonometric Ratios of
Special Angles: 30°, 45° and 60°.
2
2
1
60
45 o
1
cos 45 
2
1
o
sin 45 
2
o
tan 45  1
o
3
1
1
o
30 o
1
cos 60 
2
o
3
sin 60 
2
tan 60o  3
o
1
3
cos 30 
2
1
o
sin 30 
2
1
o
tan 30 
3
o
Trigonometric Ratios of Complementary Angles.
Q
c
b
P
R
a
b
cos(90   ) 
c
a
o
sin(90   ) 
c
a
o
tan(90   ) 
b
adjacent
a
cos 

hypotenuse
c
opposite
b
sin  

hypotenuse
c
opposite
b
tan  

adjacent
a
o
1
cos(90   )  sin , sin(90   )  cos , tan(90   ) 
tan
o
o
o