Copyright © 2009 Pearson Education, Inc.

CHAPTER 8: Applications of Trigonometry

8.1 The Law of Sines

8.2 The Law of Cosines

8.3 Complex Numbers: Trigonometric Form

8.4 Polar Coordinates and Graphs

8.5 Vectors and Applications

8.6 Vector Operations

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8.1The Law of Sines

Use the law of sines to solve triangles. Find the area of any triangle given the lengths of two

sides and the measure of the included angle.

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Solving Oblique Triangles

1. AAS: Two angles of a triangle and a side opposite one of them are known.

2. ASA: Two angles of a triangle and the included side are known.

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Solving Oblique Triangles

3. SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution,one solution,or two solutions. The latter is known as the ambiguous case.)

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Solving Oblique Triangles

4. SAS: Two sides of a triangle and the included are known.

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Solving Oblique Triangles

5. SSS: All three sides of the triangle are known.

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Law of Sines

The Law of Sines applies to the first three situations.

a

sin A=

bsinB

=c

sinC

The Law of Sines

In any triangle ABC,

A

B

C

a

b

c

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ExampleIn , e = 4.56, E = 43º, and G = 57º. Solve the triangle.

Solution:

Draw the triangle.

E =43º e=4.56F =? f =?G=57º g=?

We have AAS.

EFGΔ

Slide 8.1 - 10Copyright © 2009 Pearson Education, Inc.

ExampleSolution continued

Find F: F = 180º – (43º + 57º) = 80º

f

sin F=

esinE

Use law of sines to find the other two sides.

f

sin80º=

4.56sin43º

f =4.56sin80ºsin43º

f ≈6.58

g

sinG=

esinE

f

sin57º=

4.56sin43º

f =4.56sin57ºsin43º

g ≈5.61

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ExampleSolution continued

We have solved the triangle.

80º 6.58

5.61

43º 4.56

57º

E e

F f

G g

= == == =

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Solving Triangles SSA

When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. There are various possibilities as show in the following 8 cases.

Angle B is acuteCase 1: No solutionb < c; side b is too short to reach the base. No triangle formed.

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Solving Triangles SSA

Angle B is acuteCase 2: One solutionb < c; side b just reaches the base and is perpendicular to it.

Angle B is acuteCase 3: Two solutionsb < c; an arc of radius b meets the base at two points. This is called the ambiguous case.

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Solving Triangles SSA

Angle B is acuteCase 4: One solutionb = c; an arc of radius b meets the base at just one point other than B.

Angle B is acuteCase 5: One solutionb > c; an arc of radius b meets the base at just one point.

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Solving Triangles SSA

Angle B is obtuseCase 6: No solutionb < c; side b is too short to reach the base. No triangle formed.

Angle B is obtuseCase 7: No solutionb = c; an arc of radius b meets the base only point B. No triangle is formed.

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Solving Triangles SSA

Angle B is obtuseCase 8: One solutionb > c; an arc of radius b meets the base at just one point.

The eight cases lead us the three possibilities in the SSA situation: no solution, one solution, or two solutions.

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Example

In , b = 15, c = 20, and B = 29º. Solve the triangle.

Solution

Draw a triangle.

A =? a=?B=29º b=15C =? c=20

Find C.b

sin B=

csinC

ABCΔ

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ExampleSolution continued

There are two angles less than 180º with a sine of 0.6464: 40º and 140º. So there are two possible solutions.

15

sin29º=

20sinC

sinC =20sin29º

15≈0.6464

Possible Solution IIf C = 40º, then

A = 180º – (29º + 40º) = 111º

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These measures make a triangle as shown. Thus we have a solution.

ExampleSolution continued

Then we find a:

a

sin111º=

15sin29º

a =15sin111ºsin29º

≈29

a

sin A=

bsinB

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ExampleSolution continued

Then we find a:

a

sin111º=

15sin29º

a =15sin111ºsin29º

≈29

Possible Solution IIIf C = 140º, then

A = 180º – (29º + 140º) = 11º

a

sin A=

bsinB

These measures make a triangle as shown. Thus we have a solution.

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The Area of a Triangle

The area of any is one half the product of he lengths of two sides and the sine of the included angle:

K =12bcsinA=

12absinC =

12acsinB.

ABCΔ

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ExampleA university landscaping architecture department is designing a garden for a triangular area in a dormitory complex. Two sides of the garden, formed by the sidewalks in front of buildings A and B, measure 172 ftand 186 ft, respectively, and together form a 53º angle. The third side of the garden,formed by the sidewalk along Crossroads Avenue, measures 160 ft. What is the area of the garden to the nearest square foot?

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Example

Solution:

Use the area formula.

The area of the garden is approximately 12,775 ft2.

K =12absinC

186 ft 172 ft 53º1

sin2

K = ⋅ ⋅ ⋅

212,775 ftK ≈