Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

29
Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Transcript of Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Page 1: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Page 2: Slide 7.1 - 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 Sines

Learn the vocabulary and conventions associated with solving triangles.Learn the statement and the derivation of the Law of Sines.Learn to solve AAS and ASA triangles by using the Law of Sines. Learn to solve for possible triangles in the ambiguous SSA case.

SECTION 7.1

1

2

3

4

Page 3: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

LAW OF SINES

The following diagrams illustrate the Law of Sines.

Page 4: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

LAW OF SINES

sin A

a

sin B

b,

sin B

b

sinC

c,

sinC

c

sin A

A

sin A

a

sin B

b

sinC

c or

a

sin A

b

sin B

c

sinC

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

We can rewrite these relations in compact notation:

Page 5: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING AAS AND ASA TRIANGLES

Step 1: Find the third angle. Find the measure of the third angle by subtracting the measures of the known angles from 180º.

Step 2: Make a chart. Make a chart of the six parts of the triangle, indicating the known parts and the parts to be computed. Make a sketch of the triangle.

Page 6: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING AAS AND ASA TRIANGLES

Step 3: Apply the Law of Sines. Select two ratios of the Law of Sines in which three of the four quantities are known. Solve for the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.

Step 4: Show the solution. Show the solution by completing the chart.

Page 7: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Height of Mount Everest

From a point on a level plain at the foot of the mountain, a surveyor finds the angle of elevation of the peak of Mount Everest to be 20º. She walks 3465 meters closer (on a direct line between the first point and the base of the mountain) and finds the angle of elevation to be 23º. Estimate the height of the peak of Mount Everest to the nearest meter.

Page 8: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Height of Mount Everest

Solution

mABC 180º 23º 157º

mC 180º 20º 157º 3º

a

sin 20º

3465

sin 3º

Page 9: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

EXAMPLE 2 Height of Mount Everest

a 3465sin 20º

sin 3º22,644 meters

In triangle BCD, sin 23º h

a.

h asin 23º 22,644 sin 23º 8848 meters

Height of Mount Everest is about 8848 meters.

Page 10: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLES(AMBGUOUS CASE)

If the lengths of two sides of a triangle and the measure of the angle opposite one of theses sides are given, then depending on the measurements, there may be

1. No such triangle2. One such triangle3. Two such triangles

For this reason, Case 2 is called the ambiguous case.

Page 11: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLES(AMBGUOUS CASE)

A is an acute angle.

Page 12: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLES(AMBGUOUS CASE)

A is an acute angle.

Page 13: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLES(AMBGUOUS CASE)

A is an obtuse angle.

Page 14: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLESStep 1: Make a chart. Make a chart of the

six parts of the triangle, indicating the known parts and the parts to be computed.

Step 2: Apply the Law of Sines to the two ratios of in which three of the four quantities are known. Solve for the sine of the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.

Page 15: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLESStep 3: If the sine of the angle, say , in

Step 2 is greater than 1, there is no triangle with the given measurements. If sin in Step 2 is between 0 and 1, go to Step 4.

Step 4: Suppose the sine of the angle in Step 2 is x, with 0 ≤ x ≤ 1. If x ≠ 1, there are two possible values for the angle:(i) = sin-1

x with 0 < < 90º and (ii) = 180º –

Page 16: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

SOLVING SSA TRIANGLESStep 5: If x ≠ 1, with

(known angle) + 1 < 180º and(known angle) + 2 < 180º, thenthere are two triangles. Otherwise, there is only one triangle, and if x = 1, it is a right triangle.

Step 6: Find the third angle of the triangle(s).

Step 7: Use the Law of Sines to find the remaining side(s).

Step 8: Show solution(s).

Page 17: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3 Solving an SSA Triangle (No Solution)

Solve triangle ABC with A = 50º, a = 8 inches, and b = 15 inches.

SolutionStep 1 Make a chart.

Step 2 Apply the Law of Sines.

c = ?C = ?

b = 15B = ?

a = 8A = 50º

sin B

b

sin A

a

Page 18: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

sin B bsin A

a

EXAMPLE 3 Solving an SSA Triangle (No Solution)

sin B 15sin 50º

8

sin B 1.44

Step 3 Since sin B ≈ 1.44 > 1 and the range of the sine function is [–1, 1], we conclude that no triangle exists that is compatible with the given information.

Page 19: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 19 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5 Solving an SSA Triangle (Two Solutions)

Solve triangle ABC with A = 47º, a = 11 meters, and c = 13 meters. Round each answer to the nearest tenth.SolutionStep 1 Make a chart.

Step 2 Apply the Law of Sines.

c = 13C = ?

b = ?B = ?

a = 11A = 47º

sinC

c

sin A

a

Page 20: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 20 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

sinC csin A

a

EXAMPLE 5 Solving an SSA Triangle (Two Solutions)

sinC 13sin 47º

110.864

Step 4 Two possible values of C withsin C ≈ 0.864 areC1 ≈ sin-1

(0.864) = 59.8º andC2 ≈ 180º – 59.8º = 120.2º

Page 21: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 21 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

EXAMPLE 5 Solving an SSA Triangle (Two Solutions)

Step 5 A + C1 = 47º + 59.8º = 106.8º < 180º, A + C2 = 47º + 120.2º = 167.2º < 180º.There are two triangles with the given measurements.

Step 6 ABC1 180º 47º 59.8º 73.2º , and

ABC2 180º 47º 120.2º 12.8º .

Page 22: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 22 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

EXAMPLE 5 Solving an SSA Triangle (Two Solutions)

Step 7

b1 asinABC1

sin A

b1 11sin 73.2º

sin 47º

b1 14.4 meters

b2 asinABC2

sin A

b2 11sin12.8º

sin 47º

b1 3.3 meters

b1

sinABC1

a

sin A

b2

sinABC2

a

sin A

Page 23: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 23 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

EXAMPLE 5 Solving an SSA Triangle (Two Solutions)

Step 8 BAC1

BAC2

c = 13C2 ≈ 120.2º

b ≈ 3.3 ≈ 12.8º

a = 11A = 47ºABC2

c = 13C1 ≈ 59.8º

b1 ≈ 14.4 ≈ 73.2º

a = 11A = 47º

ABC1

Page 24: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 24 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

BEARINGSA bearing is the measure of an acute angle from due north or due south.

N 40º E N 60º W

Page 25: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 25 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

BEARINGS

S 30º W S 75º E

Page 26: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 26 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Navigating Using Bearings

A ship sailing due west at 20 miles per hour records the bearing of an oil rig at N 55.4º W. An hour and a half later the bearing of the same rig is N 66.8º E.

a. How far is the ship from the oil rig the second time?

b. How close did the ship pass to the oil rig?

Page 27: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 27 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6 Navigating Using Bearings

Solutiona. In an hour and a half the ship travels (1.5)

(20) = 30 miles due west. The oil rig (C), the starting point for the ship (A), and the position of the ship after an hour and a half (B) form the vertices of the triangle ABC.

Page 28: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 28 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

Then A = 90º – 55.4º = 34.6º and

The ship is approximately 20 miles from the oil rig when the second bearing is taken.

a

sin A

c

sinC

EXAMPLE 6 Navigating Using Bearings

a csin A

sinCa

30sin 34.6º

sin122.2º20 miles

B = 90º – 66.8º = 23.2º. ThusC = 180º – 34.6º – 23.2º = 122.2º.

Page 29: Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 7.1 - 29 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solution continued

b. The shortest distance between the ship and the oil rig is the length of the segment h in triangle ABC.

h asin B

h 20sin 23.2º

h 7.9 miles

sin B h

a

EXAMPLE 6 Navigating Using Bearings

The ship passes within 7.9 miles of the oil rig.