Additional Topics in Trigonometryteachers.dadeschools.net/rcarrasco/bzpc5e_06_02.pdf · Solving...

Chapter 6

Additional Topics

in Trigonometry

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1

6.2 The Law of Cosines


Objectives:

• Use the Law of Cosines to solve oblique triangles.

• Solve applied problems using the Law of Cosines.

• Use Heron’s formula to find the area of a triangle.


Solving Oblique Triangles

Solving an oblique triangle means finding the lengths

of its sides and the measurements of its angles.

The Law of Cosines is used to solve triangles in which

two sides and the included angle (SAS) are known, or

those in which three sides (SSS) are known.


The Law of Cosines


Solving an SAS Triangle


Example: Solving an SAS Triangle

Solve the triangle shown in the figure with A = 120°, b = 7,

and c = 8. Round lengths of sides to the nearest tenth and

angle measures to the nearest degree.

Step 1 Use the Law of Cosines to find

the side opposite the given angle. 2 2 2 2 cosa b c bc A 2 2 27 8 2(7)(8)cos120a

2 149 64 112

2a

2 169a

169 13a






Step 2 Use the Law of Sines to find the

angle opposite the shorter of the two

given sides. This angle is always acute.

sin sin

b a

B A

sin sinb A a B

sin 7sin120sin

13

b AB

a

0.4663

1sin (0.4663) 28B






Step 3 Find the third angle. Subtract

the measure of the given angle and the

angle found in step 2 from 180°.

180A B C

120 28 180C

148 180C

32C

13

28

32

a

B

C


Solving an SSS Triangle:


Example: Solving an SSS Triangle

Solve triangle ABC if a = 8, b = 10, and c = 5. Round


A

B

C

58

10

Step 1 Use the Law of Cosines to find

the angle opposite the longest side. 2 2 2 2 cosb a c ac B

2 2 22 cosac B a c b 2 2 2

cos2

a c bB

ac

2 2 28 5 10

2(8)(5)

11

80

1 11cos 82

80

Because cosB is negative, B is an obtuse angle.

180 82 98B





A

B

C

58

10

Step 2 Use the Law of Sines to find

either of the two remaining acute angles.

sin sin

b a

B A

sinsin

a BA

b

8sin98

10

0.7922

1sin (0.7922) 52





A

B

C

58

10

Step 3 Find the third angle.

Subtract the measure of the

given angle and the angle found

in step 2 from 180°.

180A B C

180C B A

180 98 52 30C

52

98

30

A

B

C


Example: Application

Two airplanes leave an airport at the same time on different

runways. One flies directly north at 400 miles per hour.

The other airplane flies on a bearing of N75°E at 350 miles

per hour. How far apart will the airplanes be after two

hours? N

S

75

A

B

C

2 400

800

2 350

700

2 2 2 2 cosb a c ac B 2 2 2700 800 2(700)(800)cos75b

2 840122.670b

840122.670 917b

After two hours, the planes are

approximately 917 miles apart.


Heron’s Formula


Example: Using Heron’s Formula

Find the area of the triangle with a = 6 meters, b = 16

meters, and c = 18 meters. Round to the nearest square

meter.

1( )

2s a b c

1(6 16 18) 20

2

Area ( )( )( )s s a s b s c

20(20 6)(20 16)(20 18)

2240 47

The area of the

triangle is

approximately

47 square meters.

Additional Topics in Trigonometryteachers.dadeschools.net/rcarrasco/bzpc5e_06_02.pdf · Solving...

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