6.16.16.16.1

The Law of SinesThe Law of Sines

Quick Review

Given / / , solve for the given variable.

1.

2.

Evaluate the expression.

8sin 323.

5Solve for the angle .

4. sin 0.2 0 90

5. sin 0.2 90 180

a b c d

b

c

x

x x

x x

Quick Review Solutions

Given / / , solve for the given variable.

1.

2.

Evaluate the expression.

8sin 323.

5Solve for the angle .

4. sin 0.2 0 90

0.848

11.53

5. sin 0.

7

162 90 180

a b c d

b

c

x

x x

ad

cad

x x

b

8.463

What you’ll learn about• Deriving the Law of Sines• Solving Triangles (AAS, ASA)• The Ambiguous Case (SSA)• Applications

… and whyThe Law of Sines is a powerful extension of

the triangle congruence theorems of Euclidean geometry.

Law of Sines

In with angles , , and opposite sides , , and , respectively,

the following equation is true:

sin sin sin.

ABC A B C a b c

A B C

a b c

Example Solving a Triangle Given Two Angles and a Side

Solve given that 38 , 46 , and 9.ABC A B a

Example Solving a Triangle Given Two Angles and a Side

Solve given that 38 , 46 , and 9.ABC A B a

Find 180 38 46 96 .

Apply the Law of Sines:

sin sin sin sin

sin38 sin 46 sin38 sin96

9 99sin 46 9sin96

sin38 sin38

10.516

C

A B A C

a b a c

b c

b c

b

14.538

The six parts of the triangle are:

38 9

46 10.516

96 14.538

c

A a

B b

C c

Example Solving a Triangle Given Two Sides and an Angle (The Ambiguous

Case)

Solve given that 7, 6, and 30 .ABC a b A

Example Solving a Triangle Given Two Sides and an

Angle (The Ambiguous Case)

Solve given that 7, 6, and 30 .ABC a b A

1

Use the Law of Sines to find .

sin30 sin

7 66sin30

sin7

25.4 or 180 25.4 154.6

If 25.4 , If

B

B

B

B B

B

154.6 ,

then 180 30 25.4 124.6 then 180 30 154.6 4.6

7sin124.6and 11.5 Since this is not possible,

sin30

B

C C

c

there is only one triangle.

The six parts of the triangle are:

30 , 7, 25.4 , 6, 124.6 , 11.5.A a B b C c

Example Finding the Height of a Pole

A road slopes 15 above the horizontal, and a vertical telephone pole

stands beside the road. The angle of elevation of the Sun is 65 , and

the pole casts a 15 foot shadow downhill along the road. Fin

d the height

of the pole.

x

15ft15º

65º

B

A

C

Example Finding the Height of a Pole

A road slopes 15 above the horizontal, and a vertical telephone pole

stands beside the road. The angle of elevation of the Sun is 65 , and

the pole casts a 15 foot shadow downhill along the road. Fin

d the height

of the pole.

x

15ft15º

65º

B

A

C

Let the height of the pole.

180 90 65 25

65 15 50

sin 25 sin50

1515sin50

27.2sin 25

The height of the pole is about 27.2 feet.

x

BAC

ACB

x

x

6.26.26.26.2

The Law of CosinesThe Law of Cosines

Quick Review

2 2 2

2 2

Find an angle between 0 and 180 that is a solution to the equation.

1. cos 4 / 5

2. cos -0.25

Solve the equation (in terms of and ) for (a) cos and

(b) , 0 180 .

3. 7 2 cos

4. 4

A

A

x y A

A A

x y xy A

y x

4 cos

5. Find a quadratic polynomial with real coefficients that has no real zeros.

x A

Quick Review Solutions

2

Find an angle between 0 and 180 that is a solution to the equation.

1. cos 4 / 5

2. cos -0.25

Solve the equation (in terms of and ) for (a) cos and

(b) , 0 180 .

3.

36.87

104.48

7

A

A

x y A

A A

2 2 2 2

-12 2

2

2 2 2 2

-12

49 49(a) (b) cos

2 2

4 4(a) (b) cos

2 cos

4. 4 4 cos

5. Find a quadratic polynomial with real coefficients that has no

4

ea

4

r

x y xy A

y x x

x y x y

xy xy

y x y x

x xA

2One answer

l ze

:

r s.

2

o

x

What you’ll learn about• Deriving the Law of Cosines• Solving Triangles (SAS, SSS)• Triangle Area and Heron’s Formula• Applications

… and whyThe Law of Cosines is an important

extension of the Pythagorean theorem, with many applications.

Law of Cosines

2 2 2

2 2 2

2 2 2

Let be any triangle with sides and angles

labeled in the usual way. Then

2 cos

2 cos

2 cos

ABC

a b c bc A

b a c ac B

c a b ab C

Example Solving a Triangle (SAS)

Solve given that 10, 4 and 25 .ABC a b C

Example Solving a Triangle (SAS)

Solve given that 10, 4 and 25 .ABC a b C

2 2 2

2

2

Use the Law of Cosines:

2 cos

16 100 2(4)(10)cos25

6.6

Use the Law of Cosines again:

10 16 43.56 2(4)(6.6)cos

cos 0.7659

140

180 140 25 15

The six parts of the triangle are:

14

c a b ab C

c

c

A

A

A

B

A

0 , 10, 15 , 4, 25 , 6.6.a B b C c

Area of a Triangle 1 1 1

Area sin sin sin2 2 2bc A ac B ab C

Heron’s Formula

Let , , and be the sides of , and let denote the

( ) / 2. Then the area of is given by

Area - .

a b c ABC s

a b c ABC

s s a s b s c

semiperimeter

Example Using Heron’s Formula

Find the area of a triangle with sides 10, 12, 14.

Example Using Heron’s Formula

Find the area of a triangle with sides 10, 12, 14.

Compute : (10 12 14) / 2 18.

Use Heron's Formula:

18 18 10 18 12 18 14

= 3456

=24 6 58.8

The area is approximately 58.8 square units.

s s

A

Chapter Test

3

2

1. Prove the identity cos3 4cos 3cos .

2. Write the expression in terms of sin and cos .

cos 2 sin 2

3. Find the general solution without using a calculator.

2cos 2 1

x x x

x x

x x

x

Chapter Test

4 2

2

4. Solve the equation graphically. Find all solutions in the interval [0,2 ).

sin 2

5. Find all solutions in the interval [0,2 ) without using a calculator.

sin 2sin 3 0

6. Solve the inequality.

x x

x x

Use any method, but give exact answers.

2cos 1 for 0 2

7. Solve , given 79 , 33 , and 7.

8. Find the area of , given 3, 5, and 6.

x x

ABC A B a

ABC a b c

Chapter Test

9. A hot-air balloon is seen over Tucson, Arizona, simultaneously by two observers at points A and B that are 1.75 mi apart on level ground and in line with the balloon. The angles of elevation are as shown here. How high above ground is the balloon?

10. A wheel of cheese in the shape of a right circular cylinder is 18 cm in diameter and 5 cm thick. If a wedge of cheese with a central angle of 15º is cut from the wheel, find the volume of the cheese wedge.

Chapter Test Solutions

2 2 3 2

3 3

3

2

cos3 cos(2 ) cos 2 cos sin 2 sin

cos sin cos 2sin cos sin cos 3cos sin

cos 3cos

1. Prove the identity cos3 4cos 3cos .

2. Write the expression in terms of sin and

1 cos 4cos 3cos .

x x x x x x x

x x x x x x x x x

x x

x x x

x x x

x

22 2

cos .

cos 2 sin 2

3. Find the general solution without using a calculator.

2co

1 4sin cos 2co

s 2 1

s sin

52 , 2

6 6

x

x x x x x

nx

x

n

Chapter Test Solutions

4 2

2

4. Solve the equation graphically. Find all solutions in the interval [0,2 ).

sin 2

5. Find all solutions in the interval [0,2 ) without using a calculator.

sin 2sin 3 0

6. Solve

1.15

3

2

x x

x

x

x

the inequality. Use any method, but give exact answers.

2cos 1 for 0 2

7. Solve

5,

3 3

6

, given 79 , 33 , and 7.

8. Find the area of ,

8 , 3.88, 6

given 3, 5,

.61

x x

ABC A B a

AB

C b

C a b

c

and 6. 7.5c

Chapter Test Solutions

9. A hot-air balloon is seen over Tucson, Arizona, simultaneously by two observers at points A and B that are 1.75 mi apart on level ground and in line with the balloon. The angles of elevation are as shown here. How high above ground is the balloon?

10. A wheel of cheese in the shape of a right circular cylinder is 18 cm in diameter and 5 cm thick. If a wedge of cheese with a central angle of 15º is cut from the wheel, find the volume of the cheese wedge.

≈0.6 mi

405π/24≈53.01