Slide 7.2 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

The Law of Cosines

Learn the statement and the derivation of the Law of Cosines.Learn to use the Law of Cosines to solve SAS triangles.Learn to use the Law of Cosines to solve SSS triangles. Learn to state and derive Heron’s formula for the area of a triangle.

SECTION 7.2

1

2

3

4

Slide 7.2 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

LAW OF COSINES

The following diagrams illustrate the Law of Cosines.

Slide 7.2 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

LAW OF COSINES

a2 b2 c2 2bccosA,

b2 c2 a2 2cacosB,

c2 a2 b2 2abcosC.

Let A, B, and C denote the measures of the angles of a triangle ABC, with opposite sides of lengths a, b, and c, respectively. Then

In words, the square of any side of a triangle is equal to the sum of the squares of the length of the other two sides, less twice the product of the lengths of the other sides and the cosine of their included angle.

Slide 7.2 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SAS TRIANGLES

Step 1: Use the appropriate form of the Law of Cosines to find the side opposite the given angle.

Step 2: Use the Law of Sines to find the angle opposite the shorter of the two given sides. Note that this angle is always an acute angle.

Step 3: Use the angle sum formula to find the third angle.

Step 4: Write the solution.

Slide 7.2 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1 Solving an SAS Triangle

Solve triangle ABC with b = 16 meters, c = 12 meters, and A = 50º. Round each answer to the nearest tenths. SolutionStep 1 Find a, the length of the side opposite

angle A. a2 b2 c2 2bccosA

a b2 c2 2bccosA

a 12.376 meters

a 16 2 12 2 2 16 12 cos 50º

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Solution continued

sinC

c

sinA

a

EXAMPLE 1 Solving an SAS Triangle

Step 2 Find C, the measure of the angle opposite the shorter of the two given sides.

sinC csinA

a

C sin 1 csinA

a

C sin 1 12sin 50º

12.376

48º

Slide 7.2 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

B180º A C

EXAMPLE 1 Solving an SAS Triangle

Step 3 Find the third angle measure, B.

B180º 50º 48º

B82ºStep 4 The solution of triangle ABC is:

c = 12 metersC ≈ 48º

b = 16 metersB ≈ 82º

a ≈ 12.4 metersA = 50º

Slide 7.2 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Using the Law of Cosines

Suppose that a Boeing 747 is flying over Disney World headed due south at 552 miles per hour. Twenty minutes later, an F-16 passes over Disney World with a bearing of N 37º E at a speed of 1250 miles per hour. Find the distance between the two planes 3 hours after the F-16 passes over Disney World. Round the answer to the nearest tenth.

Slide 7.2 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Using the Law of Cosines

1

3

SolutionSuppose the F-16 has been traveling for t hours after passing over Disney World. Then, because the Boeing 747 had a head start of 20 minutes

hour, the Boeing 747 has

been traveling

hours due south.

t 1

3

The distance between the two planes is d.

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EXAMPLE 2 Using the Law of Cosines

d 2 1250t 2 552 t 1

3

2

2 1250t 552 t 1

3

cos143º

Solution continuedUsing the Law of Cosines in triangle FDB, we have

Substitute t = 3.

d 2 28, 469,270.04

d 5335.7 miles

Slide 7.2 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSS TRIANGLES

Step 1: Use the Law of Cosines to find the side angle opposite the longest side.

Step 2: Use the Law of Sines to find either of the two remaining acute angles.

Step 3: Use the angle sum formula to find the third angle.

Step 4: Write the solution.

Slide 7.2 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3 Solving an SSS Triangle

Solve triangle ABC with a = 8, b = 5, and c = 7. Round each answer to the nearest tenths.

SolutionStep 1 Find A, the angle opposite the largest

side. a2 b2 c2 2bccosA

cosAb2 c2 a2

2bc

Acos 1 0.1429

52 72 82

257cosA0.1429

A81.8º

Slide 7.2 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

sinB

b

sinA

a

EXAMPLE 3 Solving an SSS Triangle

Step 2 Find B, using the Law of Sines.

sinBbsinA

a

Bsin 1 bsinA

a

C sin 1 5sin 81.8º

8

38.2º

Slide 7.2 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

B180º A B

Step 3 Find C by using the angle sum formula.

B180º 81.8º 38.2º

B60ºStep 4 Write the solution.

c = 7C ≈ 60º

b = 5B ≈ 38.2º

a = 8A ≈ 81.8º

EXAMPLE 3 Solving an SSS Triangle

Slide 7.2 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Solving an SSS Triangle

Solve triangle ABC with a = 2 meters, b = 9 meters, and c = 5 meters. Round each answer to the nearest tenths. SolutionStep 1 Find B, the angle opposite the longest

side. b2 c2 a2 2cacosB

cosBc2 a2 b2

2ca

52 22 92

252cosB 2.6

Range of the cosine function is [–1, 1], there is no angle B, the triangle cannot exist.

Slide 7.2 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

HERON’S FORMULA FOR SSS TRIANGLES

The area K of a triangle with sides of lengths a, b, and c is given by

where is the semiperimeter.

K s s a s b s c ,

s 1

2a b c

Slide 7.2 - 18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Using Heron’s Formula

A triangular swimming pool has side lengths 23 feet, 17 feet, and 26 feet. How many gallons of water will fill the pool to a depth of 5 feet? Round answer to the nearest whole number.

SolutionTo calculate the volume of water in the swimming pool, we first calculate the area of the triangular surface.

We have a = 23, b = 17, and c = 26.

s 1

2a b c 1

22317 26 33

Slide 7.2 - 19 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

K s s a s b s c

192.2498 5

By Heron’s formula, the area K of the triangula surface is

EXAMPLE 6 Using Heron’s Formula

K 33 33 23 33 17 33 26 K 192.2498 square feet

Volume of water in pool = surface area depth

961.25 cubic feet

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Solution continued

One cubic foot contains approximately 7.5 gallons of water.

EXAMPLE 6 Using Heron’s Formula

So 961.25 7.5 ≈ 7209 gallons of water will fill the pool.