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

Slide1

6.2

LAW OF COSINESSlide2

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 LearnSlide3

IntroductionTwo 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.Slide4

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.Slide5

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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.Slide6

Proof of the Law of Cosines

C

A

B

c

a h

b

Prove that c

2

= a

2

+ b

2

– 2ab cos C

In triangle CBD,

cos C = x / a

Then, x= a cos C (Eq #1)

Using Pythagorean Theorem

h

2

= a

2

- x2 (Eq #2)In triangle BDA,Using Pythagorean Theorem c

2 = 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

D

x

b - x

Prove:

c

2

= a

2

+ b

2

– 2ab cos CSlide7

IntroductionSlide8

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.Slide9

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

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.

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

  Slide10

Law of CosinesKnowing 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

ObtuseSlide11

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

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°.Slide12

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.

A

B

C

 Slide13

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

We will use the Law of Sines to find

B.

A

B

C

 Slide14

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

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°.Slide15

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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.Slide16

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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.

 Slide17

Example 3 – An Application of the Law of CosinesThe 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 6.13Slide18

Example 3 – Solution

In 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 thepitcher’s mound to first base is  42.43 feet.Slide19

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

Then the area of the triangle is Slide20

Example 5 – Using Heron’s Area FormulaFind 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.Slide21

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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.)Slide22

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Example 5

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

The semiperimeter

s

is

Using Heron’s formula, the area

 isSlide23

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238.2-23Four possible cases can occur when solving an oblique triangle.

When to use the Law of Sines and the Law of CosinesSlide24

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248.2-24Slide25

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