6.2 LAW OF COSINES. 2 Use the Law of Cosines to solve oblique triangles (SSS or SAS). Use the Law of...

6.2 LAW OF COSINES

2

• Use the Law of Cosines to solve oblique triangles (SSS or SAS).

• Use the Law of Cosines to model and solve real-life problems.

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

What You Should Learn

3

Introduction

Two cases remain in the list of conditions needed to solve an oblique triangle—SSS and SAS.

If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete.

In such cases, you can use the Law of Cosines.

Let's consider types of triangles with the three pieces of information shown below.

SAS

You may have a side, an angle, and then another side

AAA

You may have all three angles.

SSS

You may have all three sides

This case doesn't determine a triangle because similar triangles have the same angles and shape but "blown up" or "shrunk down"

We can't use the Law of Sines on these because we don't have an angle and a side opposite it. We need another method for SAS and SSS triangles.

Copyright © 2009 Pearson Addison-Wesley

1.1-5

8.2-5

Triangle Side Length Restriction

In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.

Proof of the Law of Cosines

C A

B

c a h

b

Prove that c2= a2 + b2 – 2ab cos CIn triangle CBD,

cos C = x / aThen, x= a cos C (Eq #1)Using Pythagorean Theorem

h2 = a2 - x2 (Eq #2)In triangle BDA,Using Pythagorean Theorem

c2 = h2 + (b – x)2

c2 = h2 + b2 – 2bx + x2 (Eq #3)Substitute with h2 Eq #2

c2 = a2 - x2 + b2 – 2bx + x2

Combine like Terms c2 = a2 - x2 + b2 – 2bx + x2

c2 = a2 + b2 – 2bx Finally, substitute x with Eq #1

c2 = a2 + b2 – 2ba cos CTherefore: c2 = a2 + b2 – 2ab cos C

Dx b - x

Prove: c2 = a2 + b2 – 2ab cos C

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Introduction

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

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



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. We use the Law of Sines to find B.

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°.

10.47 m

𝟓𝟔 .𝟎 °

𝟖𝟏 .𝟕 °

10

Law of Cosines

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.

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

Acute

Obtuse



Step 1: Use the Law of Cosines to find the side opposite the given angle.

Step 2: Use the Law of Sines to find the angle opposite the shorter of the two given sides. This angle is always acute.

Step 3: Find the third angle. Subtract the measure of the given angle and the angle found in step 2 from 180°.


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

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

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

AB

C

𝑏=15.9𝑎=9.47

𝑐=21.1

109.9 °



We will use the Law of Sines to find B.

AB

C

𝑏=15.9𝑎=9.47

𝑐=21.1

109.9 °

sin𝐵15.9

=sin 109.9

21.1

sin𝐵=15.9 ∙ sin 109.9

21.1

sin𝐵=.708 5583516

sin−1 .7085583516=45.1 °

45.1°𝐴=180 ° −10 9.9° − 45.1 °≈ 25.0 °

25.0 °



• Use the Law of Cosines to find the angle opposite the longest side.

• Use the Law of Sines to find either of the two remaining acute angles.

• Find the third angle by subtracting the measures of the angles found in steps 1 and 2 from 180°.


1.1-15

8.2-15

Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS)

A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that AC = 259 m, BC = 423 m, and angle ACB measures 132°40′. Find the distance AB.


1.1-16

8.2-16

Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS)

Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.

The distance between the two points is about 628 m.

( 𝐴𝐵 )2=( 𝐴𝐶 )2+ (𝐵𝐶 )2 −2 ( 𝐴𝐶 ) (𝐵𝐶 )cos𝐶

( 𝐴𝐵 )2=(𝟐𝟓𝟗 )𝟐+ (𝟒𝟐𝟑)𝟐−2 (𝟐𝟓𝟗 ) (𝟒𝟐𝟑 ) cos𝟏𝟑𝟐°𝟒𝟎 ′

17

Example 3 – An Application of the Law of Cosines

The pitcher’s mound on a women’s softball field is 43 feet

from home plate and the distance between the bases is 60

feet, as shown in Figure 6.13. (The pitcher’s mound is not

halfway between home plate and second base.) How far is

the pitcher’s mound from first base?

Figure 6.13

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Example 3 – SolutionIn triangle HPF, H = 45 (line HP bisects the right angle

at H), f = 43, and p = 60.

Using the Law of Cosines for this SAS case, you have

h2 = f 2 + p2 – 2fp cos H

= 432 + 602 – 2(43)(60) cos 45

1800.3.

So, the approximate distance from the

pitcher’s mound to first base is

42.43 feet.


Area Formula

• 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 c

and if the semiperimeter is

),(21

cbas

.))()(( csbsass A

Then the area of the triangle is

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Example 5 – Using Heron’s Area Formula

Find the area of a triangle having sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters.

Solution:

Because s = (a + b + c)/2

= 168/2

= 84,

Heron’s Area Formula yields

1131.89 square meters.


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8.2-21

Example 5 USING HERON’S FORMULA TO FIND AN AREA (SSS)

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 Earth.)


1.1-22

8.2-22

Example 5 USING HERON’S FORMULA TO FIND AN AREA (SSS) (continued)

The semiperimeter s is

Using Heron’s formula, the area is


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8.2-23

Four possible cases can occur when solving an oblique triangle.

When to use the Law of Sines and the Law of Cosines


1.1-24

8.2-24


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8.2-25

6.2 LAW OF COSINES. 2 Use the Law of Cosines to solve oblique triangles (SSS or SAS). Use the Law of...

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