9 6 apply the law of cosines
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9.6 Apply the Law of Cosines. In which cases can the law of cosines be used to solve a triangle? What is Heron’s Area Formula? What is the semiperimeter ?. Law of Cosines.

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9.6 Apply the Law of Cosines

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9 6 apply the law of cosines

9.6 Apply the Law of Cosines

In which cases can the law of cosines be used to solve a triangle?

What is Heron’s Area Formula?

What is the semiperimeter?


9 6 apply the law of cosines

Law of Cosines

Use the law of cosines to solve triangles when two sides and the included angle are known (SAS), or when all three side are known (SSS).


9 6 apply the law of cosines

Solve ABCwith a = 11,c = 14, and B = 34°.

b2 61.7

b2 61.7

7.85

SOLUTION

Use the law of cosines to find side length b.

b2 = a2 + c2 – 2ac cosB

Law of cosines

b2 = 112 + 142 – 2(11)(14) cos 34°

Substitute for a, c, and B.

Simplify.

Take positive square root.


9 6 apply the law of cosines

=

sin 34°

sinA

=

7.85

11

sin A

sin B

0.7836

sin A

=

a

b

A sin–1 0.7836 51.6°

11 sin 34°

The third angle Cof the triangle isC180° – 34° – 51.6° = 94.4°.

7.85

InABC,b 7.85,A 51.68,andC 94.48.

ANSWER

Use the law of sines to find the measure of angle A.

Law of sines

Substitute for a, b, and B.

Multiply each side by 11 and

Simplify.

Use inverse sine.


9 6 apply the law of cosines

SolveABC witha = 12,b = 27, and c = 20.

First find the angle opposite the longest side, AC. Use the law of cosines to solve for B.

272 = 122 + 202

cosB

=

– 2(12)(20)

– 0.3854cosB

B cos –1 (– 0.3854) 112.7°

SOLUTION

b2 = a2 + c2 – 2ac cosB

Law of cosines

272 = 122 + 202 – 2(12)(20) cosB

Substitute.

Solve for cosB.

Simplify.

Use inverse cosine.


9 6 apply the law of cosines

sin A

sin B

=

a

b

sin A

=

12

0.4100

sin A

=

sin 112.7°

12 sin 112.7°

27

The third angle Cof the triangle isC180° – 24.2° – 112.7° = 43.1°.

27

A sin–1 0.4100 24.2°

InABC,A 24.2,B 112.7,andC 43.1.

ANSWER

Now use the law of sines to find A.

Law of sines

Substitute for a, b, and B.

Multiply each side by 12 and simplify.

Use inverse sine.


9 6 apply the law of cosines

Scientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to180°, the more efficiently the organism walked.

3162 = 1552 + 1972

cosB

=

– 2(155)(197)

– 0.6062 cos B

B cos –1 (– 0.6062) 127.3°

ANSWER

The step angle Bis about 127.3°.

Science

The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.

b2 = a2 + c2 – 2ac cosB

Law of cosines

SOLUTION

3162 = 1552 + 1972 – 2(155)(197) cosB

Substitute.

Solve for cos B.

Simplify.

Use inverse cosine.


9 6 apply the law of cosines

Find the area of ABC.

b2 57

b2 57

7.55

1. a = 8,c = 10,B = 48°

SOLUTION

Use the law of cosines to find side length b.

b2 = a2 + c2 – 2ac cosB

Law of cosines

b2 = 82 + 102 – 2(8)(10) cos 48°

Substitute for a,c, and B.

Simplify.

Take positive square root.


9 6 apply the law of cosines

=

sin 48°

sinA

=

7.55

8

sin A

sin B

0.7874

sin A

=

b

a

8 sin 48°

The third angle Cof the triangle isC180° – 48° – 51.6° = 80.4°.

7.55

A sin –1 0.7836 51.6°

ANSWER

InABC,b 7.55,A 51.6°,andC 80.4°.

Use the law of sines to find the measure of angle A.

Law of sines

Substitute for a, b, and B.

Multiply each side by 8 and

simplify.

Use inverse sine.


9 6 apply the law of cosines

Find the area of ABC.

Bcos–1(– 0.0834) 85.7°

First find the angle opposite the longest side, AC. Use the law of cosines to solve for B.

a = 14,b = 16,c = 9

2.

162 = 142 + 92

cosB

=

– 2(14)(9)

– 0.0834 cosB

SOLUTION

b2 = a2 + c2 – 2ac cosB

Law of cosines

162 = 142 + 92 – 2(14)(9) cosB

Substitute.

Solve for cos B.

Simplify.

Use inverse cosine.


9 6 apply the law of cosines

sin A

sin B

a

b

The third angle Cof the triangle isC180° – 85.2° – 60.7° = 34.1°.

sin A

14

InABC,A 60.7°,B 85.2°,andC 34.1°.

ANSWER

=

sin 85.2°

=

16

A sin–1 0.8719 60.7°

14sin 85.2°

0.8719

sin A

=

16

Use the law of sines to find the measure of angle A.

Law of sines

Substitute for a,b, and B.

Multiply each side by 14 and simplify.

Use inverse sine.


Heron s area formula

Heron’s Area Formula

Heron (Hero) of Alexandria, the Greek mathematician (10 - 70 A.D.) is credited with using the law of cosines to find this formula for the area of a triangle.


9 6 apply the law of cosines

STEP 2

Use Heron’s formula to find the area of ABC.

The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown.

Area

=

18,300

=

380 (380 – 170) (380 – 240) (380 – 350)

s (s – a)(s – b)(s – c)

STEP 1

Find the semiperimeter s.

1

1

s = (a + b + c )

= (170 + 240 + 350)

2

2

Urban Planning

SOLUTION

= 380

The area of the traffic triangle is about 18,300 square yards.


9 6 apply the law of cosines

STEP 2

Use Heron’s formula to find the area of ABC.

Find the area of ABC.

Area

4.

=

18.3

=

12 (12 – 8) (12 – 11) (12 – 5)

s (s – a)(s – b)(s – c)

STEP 1

Find the semiperimeter s.

1

1

s = (a + b + c )

= (5 + 8 + 11)

2

2

SOLUTION

= 12

The area is about 18.3 square units.


9 6 apply the law of cosines

STEP 2

Use Heron’s formula to find the area of ABC.

Find the area of ABC.

Area

=

5.

13.4

=

10 (10– 4) (10 – 9) (10 – 7)

s (s – a)(s – b)(s – c)

STEP 1

Find the semiperimeter s.

1

1

2

2

s = (a + b + c )

= (4 + 9+ 7)

SOLUTION

= 10

The area is about 13.4 square units.


9 6 apply the law of cosines

STEP 2

Use Heron’s formula to find the area of ABC.

Find the area of ABC.

Area

6.

=

80.6

=

25 (25– 15) (25 – 23) (25 – 12)

s (s – a)(s – b)(s – c)

STEP 1

Find the semiperimeter s.

1

1

2

2

s = (a + b + c )

= (15 + 23+ 127)

SOLUTION

= 25

The area is about 80.6 square units.


9 6 apply the law of cosines

9.6 Assignment

Page 596, 3-39 every 3rd problem

No work is the same as a missing problem


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