There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle.
Unfortunately, the world does not consist only of right triangles…

As a matter of fact, right triangles end up being more of a rarity than commonplace.
Does that mean when we come across a situation that can only be modeled with a non-right triangle that we abandon our pursuit?….

There exists 2 Laws of Trigonometry that allow one to solve problems that involve non-right Triangles:

A
A triangle is uniquely determined by two angles and a particular side
C b
O
1 c a
O
2
B

A
If a corresponding angle and side are known, they form an
“opposing pair”
C b
O
1 c a
O
2
B

A
The Sine Law can be used to determine an unknown side or angle given an “opposing pair”
C b
O
1 c a
O
2
B

A
Find the length of b
C b
30 o c
65 o
5
B

A
Construct CN with height h
30 o b c
C
5 h
N
65 o
B

By the right triangle SIN ratio
Sin 30 o = h b
Sin 65 o = h
5
C b
A
30 o c
5
N
65 o
B

Solve both equations for h
Sin 30 o = h b
X b Sin 65 o = h
5
X 5 bSin30 o = h h = 5Sin65 o
Because the equations are equal bSin30 o = 5Sin65 o

bSin30 o = 5Sin65 o b = 5Sin65 o
Sin30 o b = 9.1
Consider the general case:

C b h
A
N c
Sin A = h b
Sin B = h bSinA = aSinB a a
B

bSinA = aSinB a a bSinA = SinB a bSinA = SinB ab b
SinA = SinB a b

Extend this to all 3 sides of a triangle, and the Sine Law is generated!
SinA = SinB a b
= SinC c

Find the length of a
C
57 o a a
Sin73 o
= 24
Sin57 o a = 27.4
A
73 o
24 c
N

Find h
5.9
O
2.9 km h
10.3
O

Find h
1. Find O
O = 180 O – 5.9
O – 10.3
O
= 163.8
O
O
5.9
O
2.9 km
10.3
O

Find X
X
X
SIN 10.3
O
=
2.9
SIN163.8
O
X = 1.86km
163.8
O
5.9
O 10.3
O
2.9 km

Find h
SIN 5.9
O = h
1.86 km h = 191.2 m
1.86 km
5.9
O
2.9 km h
10.3
O

11
48 o
Find A
A
9
SinA
11
=
Sin48
9
A = 65.3
o o
Does that make sense?

Side 9 can also be drawn as:
11
9
Could A be
65 o in this case?
48 o
A

This type of discrepancy is called the “Ambiguous Case”
Be sure to check the diagram to see which answer fits:
O, or 180 o - O

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