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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

The Law of

Sines


The law of

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…


The law of

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!!!!

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 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

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

A

C

b

5

65o

30o

B

c


Construct cn with height h

Construct CN with height h

A

C

b

5

h

30o

65o

B

c

N


By the right triangle sin ratio

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

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


The law of

bSin30o = 5Sin65o

b = 5Sin65o

Sin30o

b = 9.1

Consider the general case:


The law of

C

b

a

h

A

B

N

c

Sin A = h

Sin B = h

b

a

bSinA = aSinB


The law of

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

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

C

a

=

24

a

Sin73o

Sin57o

57o

a = 27.4

c

73o

N

A

24


The law of

h

Find h

5.9O

10.3O

2.9 km


The law of

Find h

1. Find O

O = 180O – 5.9O – 10.3O

= 163.8O

O

5.9O

10.3O

2.9 km


The law of

X

2.9

=

Find X

SIN 10.3O

SIN163.8O

X = 1.86km

163.8O

X

5.9O

10.3O

2.9 km


The law of

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

The Ambiguous Case


Find a

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

Side 9 can also be drawn as:

Could A be 65o in this case?

11

9

48o

A


The law of

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


The law of

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