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The Law of. Sines. 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.

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The Law of

Sines


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?….


No Way!!!!

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

Law of Sines

Law of Cosines


A triangle is uniquely determined by two angles and a particular side

A

C

b

a

O2

O1

B

c


If a corresponding angle and side are known, they form an “opposing pair”

A

C

b

a

O2

O1

B

c


The Sine Law can be used to determine an unknown side or angle given an “opposing pair”

A

C

b

a

O2

O1

B

c


Find the length of b

A

C

b

5

65o

30o

B

c


Construct CN with height h

A

C

b

5

h

30o

65o

B

c

N


By the right triangle SIN ratio

Sin 30o = h

Sin 65o = h

b

5

A

C

b

5

h

h

30o

30o

65o

65o

B

c

N


Solve both equations for h

Sin 30o = h

Sin 65o = h

X 5

X b

b

5

bSin30o = h h = 5Sin65o

Because the equations are equal

bSin30o = 5Sin65o


bSin30o = 5Sin65o

b = 5Sin65o

Sin30o

b = 9.1

Consider the general case:


C

b

a

h

A

B

N

c

Sin A = h

Sin B = h

b

a

bSinA = aSinB


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

= SinC

a

b

c


Find the length of a

C

a

=

24

a

Sin73o

Sin57o

57o

a = 27.4

c

73o

N

A

24


h

Find h

5.9O

10.3O

2.9 km


Find h

1. Find O

O = 180O – 5.9O – 10.3O

= 163.8O

O

5.9O

10.3O

2.9 km


X

2.9

=

Find X

SIN 10.3O

SIN163.8O

X = 1.86km

163.8O

X

5.9O

10.3O

2.9 km


h

SIN 5.9O = h

Find h

1.86 km

h = 191.2 m

1.86 km

5.9O

10.3O

2.9 km


The Ambiguous Case


Find A

SinA

Sin48o

11

=

11

9

9

A

=

65.3o

Does that make sense?

48o

A

No Way!!!


Side 9 can also be drawn as:

Could A be 65o in this case?

11

9

48o

A


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


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