8-6 Law of Sines and Cosines (Presentation)

8/7/2019 8-6 Law of Sines and Cosines (Presentation)

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8-6 Laws of Sines and Cosines

Unit 8 Trigonometric and Circular Functions


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Concepts and ObjectivesLaw of Sines and Cosines (Obj. #31)

Use the law of sines to solve a triangle for a missingside or angle.Resolve ambiguous cases of the law of sines.

n t e area o a tr ang e us ng s ne.Use the law of cosines to solve a triangle for a missingside or angle.Use Herons formula to find the area of a triangle.


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Beyond Right Angle TrigonometryWhen we first started talking about trigonometry, we

started off with right triangles and defined the trigfunctions as ratios between the various sides.

,such that if any three of the six side and angle measuresof a triangle are known (including at least one side), thenthe other three can be found.


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Law of Sines

In any triangle ABC , with sides a , b, and c,

= =a b c

A

c b

(i.e. the lengths of the sides in a triangle areproportional to the sines of the measures of the

angles opposite them).

C B a


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Law of Sines

To solve for an angle, we can also write this as

= =sin sin sin A B C

A

c b

C B a

When using the law of sines, a good strategy is to select an equation so that the unknown variable is in thenumerator and all other variables are known.


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Law of SinesExample: Solve triangle BUGif B = 32.0, U = 81.8, and

b = 42.9 cm. G

=sin sin

b uB U

U B

g

u 42.9 cm

32.0 81.8

=

42.9

sin32.0 sin81.8u

=

42.9sin81.8

sin32.0u

80.1 cm


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Law of SinesExample: Solve triangle BUGif B = 32.0, U = 81.8, and

b = 42.9 cm. G

To find G, recall that the anglesof a triangle add up to 180.

U B

g

. cm 42.9 cm

32.0 81.8

ere ore, m = . .

Now that we have G, we canfind g:

=

42.9sin32.0 sin66.2

g

74.1 cm g


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Law of SinesIf we are given the lengths of two sides and the angle

opposite one of them, then zero, one, or two suchtriangles may exist. This situation is called theambiguous case of the law of sines.


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Law of SinesExample: Solve triangle PIE if I = 55, i = 8.94 m, and

p = 25.1 m.=

sin sinP I p i

Since sin P cannot be greater than 1, there can be nosuch angle.

=s n s n

25.1 8.94

=

25.1sin55sin

8.94P

sin 2.29986P


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Law of SinesApplying the law of sines:

For any angle of a triangle, 0 < sin 1. (If sin = 1,then = 90 and the triangle is a right triangle.)sin = sin (180 ) (Supplementary angles have the

.The smallest angle is opposite the shortest side, thelargest angle is opposite the longest side, and themiddle-valued angle is opposite the intermediate

side.


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Law of SinesExample: Solve triangle ABC if A = 55.3, a = 22.8 ft, and

b = 24.9 ft.

Use the law of sines to find angle B:

=s n . s n

22.8 24.9

=

24.9sin55.3sin

22.8B

sin .8978678B


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Law of SinesExample: Solve triangle ABC if A = 55.3, a = 22.8 ft, and

b = 24.9 ft.sin B .8978678

There are two angles B between 0 and 180 that satisfy.

sin 1 (.8978678 ) 63.9Supplementary angles have the same since value, soanother possible value of B is

180 63.9 = 116.1Since A + B < 180, both values will work.


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Law of CosinesIn any triangle ABC , with sides a , b, and c,

a 2 = b2 + c2 2 bc cos Ab2 = a 2 + c2 2 ac cos B

A

c b

c = a a cosC B a


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Area FormulasIf a triangle has sides of lengths a , b, and c, the area is

given by the following formula:= = =

1 1 1sin or sin or sin

2 2 2bc A ab C ac B A A A

The semiperimeter, s, of a triangle is

and the area of a triangle (Herons formula) is thus:

( )= + +1

2s a b c

( )( )( )= s s a s b s c A


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Homework College Algebra (brown book)

Page 742: 10-60 ( 5s), 75, 80, 85Turn in: 20, 40, 60, 80Page 755: 15-60 ( 5s), 65, 70, 75

Turn in: 20, 30, 50, 70

Classwork: Algebra & Trigonometry (green book)

Page 887: 1, 3, 5Page 879: 1, 3, 5Page 883: 1, 3, 5

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8-6 Law of Sines and Cosines (Presentation)