45 – 45 – 90 Triangle:
45⁰
The hypotenuse is
2 times the hypotenuse.
30 – 60 – 90 Triangle:
i) The hypotenuse is twice the
shorter leg.
ii)
The longer leg is 3 times the shorter leg.
60 ⁰
TRIANGLE TRIGONOMETRY
I) Right Triangle Trigonometry
B
Hypotenuse
Side adjacent to B
C
A
Side opposite B
B
Pythagorean Theorem
In any right triangle,
c
a
c2 = a2 + b2
A
b
C
B
c
a
A
C
b
TRIGONOMETRIC RATIOS
length of side oposite angle A a
sin A =

length of hypotenuse
c
length of side adjacent to angle A b
cos A 

length of hypotenuse
c
length of side opposite angle A
a
tan A 

length of side adjacent to angle A b
Example 1: Find the three trigonometric ratios for
angle A in the triangle.
B
144 m
C
156 m
60.0 m
sin A =
length of side oposite angle A a 144 m
 
 0.9231
length of hypotenuse
c 156 m
cos A 
length of side adjacent to angle A b 60.0 m
 
 0.3846
length of hypotenuse
c 156 m
A
length of side opposite angle A
a
tan A 

length of side adjacent to angle A b

144 m
= 2.400
60 m
We can find the measures of angles A, B, and C by
using a trig table or the second function on a
calculator.
We know:
A + B + C = 1800, and C = 900
therefore, A + B = 900
144 m
A  sin (
)  67.4
156 m
1
B = 900 – 67.40 = 22.60
Mnemonic device for remembering sine,
cosine, and tangents ratios:
SOHCAHTOA
Opposite
Sine 
Hypotenuse
Opposite
Tangent =
Adjacent
Adjacent
Cosine =
Hypotenuse
SOLVING RIGHT TRIANGLES
To solve a triangle means to find the measures of the various
parts of a triangle that are not given.
Examples: Draw, label, then solve the right triangle
described below. C is the right angle.
1) B = 36.70, b = 19.2m
A
c
b
C
a
B
A = 53.30, c = 32.1m, and a = 25.8 m
2) Find AC in right triangle ABC. a = 225m, and
A = 10.10 . C is the right angle. (Don’t use the
Pythagorean Theorem.)
B
225 m
tan 10.1 
b
0
225 m
C
b(tan 10.1 )  225 m
0
10.10
A
225 m
b=
0
tan 10.1
b = 1260 m
Some Useful Trigonometric Relationships
b
c
a
90o
a
For Right Triangles:
b
c 2 = a 2 + b 2 PYTHAGOREAN THEOREM
a
b
a
sinα =
cosα =
tanα =
c
c
b
b
a
b
sinβ =
cosβ =
tanβ =
c
c
a
Engineering Fundamentals, By Saeed Moaveni, Third Edition,
Copyrighted 2007
7-10
The sine and cosine law (rule) for any
arbitrarily shaped triangle:
A
b
c
a
b
c
law of sine :
=
=
sinA sinB sinC
C
a
B
In order to use the law of sines, you must know
a) two angles and any side, or
b) two sides and a angle opposite one of them.
The law of sines:
a) two angles and any side, or
b) two sides and a angle opposite
one of them.
Example 1: Solve the triangle if C = 28.00, c = 46.8cm,
and B = 101.50
A
c
b

sin C sin B
46.8cm
b

0
sin 28.0
sin101.50
b(sin28.00 )  (sin 101.50 )(46.8 cm)
B
C
(sin 101.50 )(46.8 cm)
b
 97.7 cm
0
sin 28.0
A = 1800 – B – C = 1800 -101.50 – 28.00 = 50.50
To find side a,
c
a

sin C sin A
46.8cm
a

0
sin28.0 sin50.50
a(sin28.0 )  (sin50.5 )(46.8cm)
0
0
(sin50.50 )(46.8cm)
a
(sin28.00 )
 76.90
The solution is a = 76.9 cm, b = 97.7 cm, and A = 50.5
law of cosine : a 2 = b 2 + c 2 - 2bccosA
b 2 = a 2 + c 2 - 2accosB
A
b
c 2 = a 2 + b 2 - 2abcosC
c
C
B
a
In order to use the law of
cosines you must know
C
A
B
a) two sides and the
included angle, or
b) all three sides.
Example 2: Solve the triangle if A = 115.20, b = 18.5 m,
and c = 21.7m.
A
To find side a,
a2 = b2 + c2 – 2bc cos A
a
b= 18.5m
A
115.20
B
c = 21.7 m
a2 = (18.5 m)2 + (21.7 m)2 – 2(18.5 m)(21.7 m)(cos 115.20)
a = 34.0 m
To find angle B, use the law of sines (it requires
less computation.)
B = 29.50, C = 35.30, a = 34.0 m
Length and Coordinate Systems
Coordinate Systems
are used to locate
things with respect to
a known origin
Types of coordinate
systems:
• Cartesian
• cylindrical
• spherical
Engineering Fundamentals, By Saeed Moaveni,
Third Edition, Copyrighted 2007
7-17