Copyright © 2007 Pearson Education, Inc. Slide 10-2

Chapter 10: Applications of Trigonometry; Vectors

10.1 The Law of Sines

10.2 The Law of Cosines and Area Formulas

10.3 Vectors and Their Applications

10.4 Trigonometric (Polar) Form of Complex Numbers

10.5 Powers and Roots of Complex Numbers

10.6 Polar Equations and Graphs

10.7 More Parametric Equations

Copyright © 2007 Pearson Education, Inc. Slide 10-3

10.2 The Law of Cosines and Area Formulas

• SAS or SSS forms a unique triangle

• Triangle Side Restriction– In any triangle, the sum of the lengths of any two

sides must be greater than the length of the remaining side.

Copyright © 2007 Pearson Education, Inc. Slide 10-4

10.2 Derivation of the Law of Cosines

Let ABC be any oblique triangle drawn with its vertices labeled as in the figure below.

cx

Bcy

B cosandsin

BcxBcy cosandsin The coordinates of point A become (c cos B, c sin B).

Copyright © 2007 Pearson Education, Inc. Slide 10-5

10.2 Derivation of the Law of Cosines

Point C has coordinates (a, 0) and AC has length b.

22 )0sin()cos( BcaBcb

Baccab

BacBBca

BcaBacBcb

cos2

cos2)cos(sin

sincos2cos

222

2222

222222

This result is one form of the law of cosines. Placing A or C at the origin would have given the same result, but with the variables rearranged.

Copyright © 2007 Pearson Education, Inc. Slide 10-6

10.2 The Law of Cosines

The Law of Cosines

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

.cos2

cos2

cos2

222

222

222

Cabbac

Baccab

Abccba

Copyright © 2007 Pearson Education, Inc. Slide 10-7

10.2 Using the Law of Cosines to Solve a Triangle (SAS)

Example Solve triangle ABC if

A = 42.3°, b = 12.9 meters, and

c = 15.4 meters.

Abccba cos2222 3.42cos)4.15)(9.12(24.159.12 222 a

7.1092 ameters47.10a

Copyright © 2007 Pearson Education, Inc. Slide 10-8

10.2 Using the Law of Cosines to Solve a Triangle (SAS)

B must be the smaller of the two remaining angles since it is opposite the shorter of the two sides b and c. Therefore, it cannot be obtuse.

9.12

sin

47.10

3.42sin B

47.10

3.42sin9.12sin

B

7.81180 BAC

0.56B

Caution If we had chosen to find C rather than B, we would not have known whether C equals 81.7° or its supplement, 98.3°.

Copyright © 2007 Pearson Education, Inc. Slide 10-9

10.2 Using the Law of Cosines to Solve a Triangle (SSS)

Example Solve triangle ABC if a = 9.47 feet, b =15.9 feet, and c = 21.1 feet.

Solution We solve for C, the largest angle, first. If cos C < 0, then C will be obtuse.

Cabbac cos2222

ab

cbaC

2cos

222

)9.15)(47.9(2

)1.21()9.15()47.9( 222

34109402.9.109C

Copyright © 2007 Pearson Education, Inc. Slide 10-10

10.2 Using the Law of Cosines to Solve a Triangle (SSS)

Verify with either the law of sines or the law of cosinesthat B 45.1°. Then,

.0.25

9.1091.45180

180

CBA

Copyright © 2007 Pearson Education, Inc. Slide 10-11

10.2 Summary of Cases with Suggested Procedures

Oblique Triangle Suggested Procedure for Solving

Case 1: SAA or ASA 1. Find the remaining angle using the angle sum formula (A + B + C = 180°).

2. Find the remaining sides using the law of sines.

Oblique Triangle Suggested Procedure for Solving

Case 2: SSA This is the ambiguous case; 0, 1, or 2 triangles.

1. Find an angle using the law of sines.

2. Find the remaining angle using the angle sum formula.

3. Find the remaining side using the law of sines.

If two triangles exist, repeat steps 2 and 3.

Copyright © 2007 Pearson Education, Inc. Slide 10-12

10.2 Summary of Cases with Suggested Procedures

Oblique Triangle Suggested Procedure for Solving

Case 3: SAS 1. Find the third side using the law of cosines.

2. Find the smaller of the two remaining angles using the law of sines.

3. Find the remaining angle using the angle sum formula.

Oblique Triangle Suggested Procedure for Solving

Case 4: SSS 1. Find the largest angle using the law of cosines.

2. Find either remaining angle using the law of sines.

3. Find the remaining angle using the angle sum formula.

Copyright © 2007 Pearson Education, Inc. Slide 10-13

10.2 Area Formulas

• The law of cosines can be used to derive a formula for the area of a triangle given the lengths of three sides known as Heron’s Formula.

Heron’s Formula

If a triangle has sides of lengths a, b, and cand if the semiperimeter is

),(21

cbas

.))()(( csbsass A

Then the area of the triangle is

Copyright © 2007 Pearson Education, Inc. Slide 10-14

10.2 Using Heron’s Formula to Find an Area

Example The distance “as the crow flies” from Los Angeles

to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of the earth.)

Solution

))()(( csbsass A

5.2604)24273312451(2

1s

)24275.2604)(3315.2604)(24515.2604(5.2604

700,401

Copyright © 2007 Pearson Education, Inc. Slide 10-15

Area of a Triangle

In any triangle ABC, the area A is given by any of the following:

10.2 Area of a Triangle Given SAS

• The area of any triangle is given by A = ½bh, where b is its base and h is its height.

Achch

A sinorsin

Abcbh sin21

21

Area A

BacCabAbc sin21

and,sin21

,sin21

AAA

Copyright © 2007 Pearson Education, Inc. Slide 10-16

10.2 Finding the Area of a Triangle (SAS)

Example Find the area of triangle ABC in the figure.

Solution We are given B = 55°, a = 34 feet, and c = 42 feet.

Bac sin2

1A

55sin)42)(34(2

1

feet square 585