6.1 law of sines 6.2 law of cosines

24
6.1 LAW OF SINES Copyright © Cengage Learning. All rights reserved. 6.2 LAW OF COSINES

Upload: others

Post on 12-Feb-2022

26 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: 6.1 LAW OF SINES 6.2 LAW OF COSINES

6.1 LAW OF SINES

Copyright © Cengage Learning. All rights reserved.

6.2 LAW OF COSINES

Page 2: 6.1 LAW OF SINES 6.2 LAW OF COSINES

2

Introduction

Page 3: 6.1 LAW OF SINES 6.2 LAW OF COSINES

3

Introduction

Oblique triangles—triangles that have no right angles.

Law of Sine

1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them

(SSA)

Law of Cosine

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

Page 4: 6.1 LAW OF SINES 6.2 LAW OF COSINES

4

Introduction

Page 5: 6.1 LAW OF SINES 6.2 LAW OF COSINES

5

Introduction

Page 6: 6.1 LAW OF SINES 6.2 LAW OF COSINES

6

Example 1 – Given Two Angles and One Side—AAS

For the triangle in figure, C = 120, B = 29, andb = 28 feet.

Find the remaining angle and sides.

Solution:

The third angle of the triangle is

A = 180 – B – C

= 180 – 29 – 102

= 49. .

Page 7: 6.1 LAW OF SINES 6.2 LAW OF COSINES

7

Example 1 – Solution

The third angle of the triangle is

A = 180 – B – C

= 180 – 29 – 102

= 49.

By the Law of Sines, you have

.

Page 8: 6.1 LAW OF SINES 6.2 LAW OF COSINES

8

The Ambiguous Case (SSA)

Page 9: 6.1 LAW OF SINES 6.2 LAW OF COSINES

9

The Ambiguous Case (SSA)

Page 10: 6.1 LAW OF SINES 6.2 LAW OF COSINES

10

Example 3 – Single-Solution Case—SSA

For the triangle in Figure 6.4, a = 22 inches, b = 12 inches,

and A = 42. Find the remaining side and angles.

One solution: a b

Figure 6.4

Page 11: 6.1 LAW OF SINES 6.2 LAW OF COSINES

11

Example 3 – Solution

By the Law of Sines, you have

Reciprocal form

Multiply each side by b.

Substitute for A, a, and b.

B is acute.

Page 12: 6.1 LAW OF SINES 6.2 LAW OF COSINES

12

Example 3 – Solution

Now, you can determine that

C 180 – 42 – 21.41

= 116.59.

Then, the remaining side is

cont’d

Page 13: 6.1 LAW OF SINES 6.2 LAW OF COSINES

13

Area of an Oblique Triangle

Page 14: 6.1 LAW OF SINES 6.2 LAW OF COSINES

14

Area of an Oblique Triangle

Area = (base)(height) = (c)(b sin A) = bc sin A.

A is acute. A is obtuse.

Page 15: 6.1 LAW OF SINES 6.2 LAW OF COSINES

15

Area of an Oblique Triangle

Page 16: 6.1 LAW OF SINES 6.2 LAW OF COSINES

16

Example 6 – Finding the Area of a Triangular Lot

Find the area of a triangular lot having two sides of lengths

90 meters and 52 meters and an included angle of 102.

Solution:

Consider a = 90 meters, b = 52 meters, and angle

C = 102, as shown in figure.

Then, the area of the triangle is

Area = ab sin C

= (90)(52)(sin 102)

2289 square meters.

Page 17: 6.1 LAW OF SINES 6.2 LAW OF COSINES

17

Example 1 – Three Sides of a Triangle—SSS

Find the three angles of the triangle in Figure 6.11.

Solution:

It is a good idea first to find the angle opposite the longest

side—side b in this case. Using the alternative form of the

Law of Cosines, you find that

Figure 6.11

Page 18: 6.1 LAW OF SINES 6.2 LAW OF COSINES

18

Example 1 – Solution

Because cos B is negative, you know that B is an obtuse

angle given by B 116.80.

At this point, it is simpler to use the Law of Sines to

determine A.

cont’d

Page 19: 6.1 LAW OF SINES 6.2 LAW OF COSINES

19

Example 1 – Solution

You know that A must be acute because B is obtuse, and

a triangle can have, at most, one obtuse angle.

So, A 22.08 and

C 180 – 22.08 – 116.80

= 41.12.

cont’d

Page 20: 6.1 LAW OF SINES 6.2 LAW OF COSINES

20

Introduction

Do you see why it was wise to find the largest angle first in

Example 1? Knowing the cosine of an angle, you can

determine whether the angle is acute or obtuse. That is,

cos > 0 for 0 < < 90

cos < 0 for 90 < < 180.

So, in Example 1, once you found that angle B was obtuse,

you knew that angles A and C were both acute.

If the largest angle is acute, the remaining two angles are

acute also.

Acute

Obtuse

Page 21: 6.1 LAW OF SINES 6.2 LAW OF COSINES

21

Application

Page 22: 6.1 LAW OF SINES 6.2 LAW OF COSINES

22

Example 7 – An Application of the Law of Sines

The course for a boat race starts at point A in Figure 6.9

and proceeds in the direction S 52 W to point B, then in

the direction S 40 E to point C, and finally back to A. Point

C lies 8 kilometers directly south of point A. Approximate

the total distance of the race course.

Figure 6.9

Page 23: 6.1 LAW OF SINES 6.2 LAW OF COSINES

23

Example 7 – Solution

Because lines BD and AC are parallel, it follows that

BCA CBD.

Consequently, triangle ABC has the measures shown in

Figure 6.10.

The measure of angle B is

180 – 52 – 40 = 88.

Using the Law of Sines,

Figure 6.10

Page 24: 6.1 LAW OF SINES 6.2 LAW OF COSINES

24

Example 7 – Solution

Because b = 8,

and

The total length of the course is approximately

Length 8 + 6.308 + 5.145

=19.453 kilometers.

cont’d