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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
No Way!!!! rarity than commonplace.

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

Law of Sines

Law of Cosines



If a corresponding angle and side are known they form an opposing pair
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
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
Find the length of b angle given an “opposing pair”

A

C

b

5

65o

30o

B

c


Construct cn with height h
Construct CN with height h angle given an “opposing pair”

A

C

b

5

h

30o

65o

B

c

N


By the right triangle sin ratio
By the right triangle SIN ratio angle given an “opposing pair”

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
Solve both equations for h angle given an “opposing pair”

Sin 30o = h

Sin 65o = h

X 5

X b

b

5

bSin30o = h h = 5Sin65o

Because the equations are equal

bSin30o = 5Sin65o


bSin30 angle given an “opposing pair”o = 5Sin65o

b = 5Sin65o

Sin30o

b = 9.1

Consider the general case:


C angle given an “opposing pair”

b

a

h

A

B

N

c

Sin A = h

Sin B = h

b

a

bSinA = aSinB


bSinA = aSinB angle given an “opposing pair”

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
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
Find the length of a is generated!

C

a

=

24

a

Sin73o

Sin57o

57o

a = 27.4

c

73o

N

A

24


h is generated!

Find h

5.9O

10.3O

2.9 km


Find h is generated!

1. Find O

O = 180O – 5.9O – 10.3O

= 163.8O

O

5.9O

10.3O

2.9 km


X is generated!

2.9

=

Find X

SIN 10.3O

SIN163.8O

X = 1.86km

163.8O

X

5.9O

10.3O

2.9 km


h is generated!

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
The Ambiguous Case is generated!


Find a
Find A is generated!

SinA

Sin48o

11

=

11

9

9

A

=

65.3o

Does that make sense?

48o

A

No Way!!!


Side 9 can also be drawn as
Side 9 can also be drawn as: is generated!

Could A be 65o in this case?

11

9

48o

A


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


Page 366 is generated!


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