chapter 6 additional topics in trigonometry copyright © 2014, 2010, 2007 pearson education, inc. 1...

Chapter 6Additional Topicsin Trigonometry

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

6.1 The Law of Sines


Objectives:

• Use the Law of Sines to solve oblique triangles.• Use the Law of Sines to solve, if possible, the

triangle or triangles in the ambiguous case.• Find the area of an oblique triangle using the sine

function.• Solve applied problems using the Law of Sines.


Oblique Triangles

An oblique triangle is a triangle that does not contain a right angle. An oblique triangle haseither three acute angles

or two acute angles andone obtuse angle.

The relationships among the sidesand angles of right triangles defined by the trigonometric functions are not valid for oblique triangles.


The Law of Sines


Solving Oblique Triangles

Solving an oblique triangle means finding the lengths of its sides and the measurements of its angles.

The Law of Sines can be used to solve a triangle in which one side and two angles are known. The three known measurements can be abbreviated using SAA (a side and two angles are known) or ASA (two angles and the side between them are known).


Example: Solving an SAA Triangle Using the Law of Sines

Solve the triangle with A = 64°, C = 82°, and c = 14 centimeters. Round lengths of sides to the nearest tenth.

180A B C 64 82 180B

146 180B

34B

sin sina cA C

sin sina C c A

sinsinc A

aC

14sin 64sin82

12.7 cm


Example: Solving an SAA Triangle Using the Law of Sines (continued)

Solve the triangle with A = 64°, C = 82°, and c = 14 centimeters. Round lengths of sides to the nearest tenth.

sin sinb cB C

sin sinb C c B

sinsinc B

bC

14sin34sin82

7.4 cm

34

12.7 cm

7.4 cm

B

a

b


Example: Solving an ASA Triangle Using the Law of Sines

Solve triangle ABC if A = 40°, C = 22.5°, and b = 12. Round measures to the nearest tenth.

A

B

C12

180A B C 40 22.5 180B

62.5 180B 117.5B

sin sinb aB A

sin sinb A a B

sinsinb A

aB

12sin 40sin117.5

8.7


Example: Solving an ASA Triangle Using the Law of Sines (continued)

Solve triangle ABC if A = 40°, C = 22.5°, and b = 12. Round measures to the nearest tenth.

A

B

C12

sin sinb cB C

sin sinc B b C

sinsinb C

cB

12sin 22.5sin117.5

5.2

117.5

8.7

5.2

B

a

c


The Ambiguous Case (SSA)

If we are given two sides and an angle opposite one of the two sides (SSA), the given information may result in one triangle, two triangles, or no triangle at all.

SSA is known as the ambiguous case when using the Law of Sines because the given information may result in one triangle, two triangles, or no triangle at all.


The Ambiguous Case (SSA) (continued)


Example: Solving an SSA Triangle Using the Law of Sines (No Solution)

Solve triangle ABC if A = 50°, a = 10, and b = 20.

sin sina bA B

sin sina B b A

sinsin

b AB

a 20sin50

10 1.53

There is no angle B for which the sine is greater than 1.There is no triangle with the given measurements.


Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions)

Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree.

sin sina bA B

sin sina B b Asin

sinb A

Ba

16sin3512

0.7648

1sin 0.7648 50 There are two angles between 0° and180° for which sinB = 0.7648


Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) (continued)


1sin 0.7648 50

1 50B

2 180 50 130B

1 35 50 85A B

2 35 130 165A B

there are two possible solutions.1 2180 and 180 ,A B A B




1 50B 2 130B

1 1 180A B C 2 2 180A B C

135 50 180C

185 180C

1 95C

235 130 180C

2165 180C

2 15C




1 50B 1 95C 2 130B 2 15C

1

1 1sin sinb cB C

2

2 2sin sinb cB C

1 1 1sin sinb C c B2 2 2sin sinb C c B

11

1

sinsinb C

cB

22

2

sinsinb C

cB

16sin95

sin50

20.8

16sin15sin130

5.4




There are two triangles. In one triangle, the solution is

In the other triangle, the solution is

1 50B 1 95C 1 20.8c

2 130B 2 15C 2 5.4c


The Area of an Oblique Triangle

The area of a triangle equals one-half the product of the lengths of two sides times the sine of their included angle. In the figure, this wording can be expressed by the formulas

1 1 1Area sin sin sin

2 2 2bc A ab C ac B


Example: Finding the Area of an Oblique Triangle

Find the area of a triangle having two sides of length 8 meters and 12 meters and an included angle of 135°. Round to the nearest square meter.

8 m 12 m135

A B

C 1Area sin

2ab C

1(12)(8)(sin135 )

2

34 sq m


Example: Application

Two fire-lookout stations are 13 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is N35°E and the bearing of the fire from station B is N49°W. How far, to the nearest tenth of a mile, is the fire from station B?

A B

C

35 49

13

90 35 55A 90 49 41B

180A B C 55 41 180C

96 180C 84C


Example: Application (continued)

Two fire-lookout stations are 13 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is N35°E and the bearing of the fire from station B is N49°W. How far, to the nearest tenth of a mile, is the fire from station B?

A B

C

35 49

sin sinc aC A

sin sinc A a Csin

sinc A

aC

13sin55sin84

11

The fire is approximately 11 miles from station B.

chapter 6 additional topics in trigonometry copyright © 2014, 2010, 2007 pearson education, inc. 1...

Documents