Geometry Law of Sines and Law of Cosines

CONFIDENTIAL 1

GeometryGeometry

Law of Sines and Law of Sines and Law of CosinesLaw of Cosines

CONFIDENTIAL 2

Warm up

Use the given trigonometric ratio to determine which angle of the triangle is /A.

2

1 17 cm

15 cm

8 cma) cos A = 15/17

b) sin A = 15/17

c) tan A = 1.875

1) 22) 13) 1

CONFIDENTIAL 3

Law of Sines and Law of Cosines

In this lesson, you will learn to solve any triangle. To do so, you will need to

calculate trigonometric ratios for angle measures up to 180˚.

CONFIDENTIAL 4

Finding Trigonometric Ratios for Obtuse Angles

Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.

A) sin 135˚ B) tan 98 ˚ C) cos 108 ˚

sin 135˚ = 0.71sin 135˚ = 0.71 tan 98tan 98˚= -0.72 cos cos 108˚=-0.31

CONFIDENTIAL 5

Now you try!

1) Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.

a) tan 175˚ b) cos 92˚ c) sin 160˚

1a) -0.091b) -0.031c) 0.34

CONFIDENTIAL 6

h

c

b a

C

BA

You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.

In ∆ABC, let h represent the length of the altitude from C to AB.

From the diagram, sin A = (h), and sin B = (h).

By solving for h, you find that h = bsin A

And h = a.sinB .So b sin A = a.sinB,

And (sin A/a) = (sin B/b).

You can use another altitude to show that these ratios

Equal (sin C/c).

‾b a

CONFIDENTIAL 7

The Law of Sines

Theorem 1.1

For any ABC with side lengths a, b, and c.

sin A

a =

sin B

b =

sin C

c.

ab

cA B

C

CONFIDENTIAL 8

You can use the Law of Sines to solve a triangle if you are given

• two angle measures and any side length (ASA or AAS) or

• two side length and a non-include angle measure (SSA).

CONFIDENTIAL 9

Use the Law of Sines

Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.

32

105

F

D

E

Law of Sines

A) DF

sin D

EF =

sin E

DF

sin 105

18 =

sin 32

DF DFsin 105 18sin32

DF = 18sin 32

sin 105 9.9

Substitute the given values.

Cross Products Property

Divide both sides by sin 105˚.

Next page

CONFIDENTIAL 10

B) mS

sin T

RS =

sin S

RT

sin 75

7 =

sin S

5

sin S = 5 sin 75

7

mS sin-15 sin 75

7 44

75

5

7

S

T

R

Law of Sines


Multiply both sides by 5.

Use the inverse sine function to find m/s.

CONFIDENTIAL 11

Now you try!

2) Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree.

L

18

7.6

4.3

10

622

67

125

4450

39

88

J

KPN

M

Y

ZX

C

B

AmX

AC

MLNP

d)c)

b)a)

2a) 22b) 1

CONFIDENTIAL 12

The Law of Sine cannot be used to solve every triangle. If you know two side lengths and the include angle measure or if you know all three side lengths. You cannot use the Law of Sines.

Instead, you can apply the Law of Cosines.

CONFIDENTIAL 13

Theorem 1.2

The Law of Cosines

C

BAc

b a

For any ABC with side lengths a, b, and c: a2 = b2 + c2 - 2bc cosA b2 = a2 + c2 - 2ac cosB

c2 = a2 + b2 - 2ab cosC

CONFIDENTIAL 14

You can use the Law of Cosines to solve a triangle if you are given

• two side lengths and the included angle measure (SAS) or

• three side lengths (SSS).

CONFIDENTIAL 15

Using the Law of CosinesFind each measure. Round lengths to the nearest tenth

and angle measures to the nearest degree.

9

14

62

C

B

A

Law of Cosines


Simplify.

A ) BC

BC2 = AB2 + AC2 - 2(AB)(AC)cos A = 142 + 92 - 2(14)(9)cos 62

BC2 158.6932

BC 12.6 Find the square root of both sides.

Next page

CONFIDENTIAL 16

B ) mR

ST2 = RS2 + RT2 - 2(RS)(RT)cos R

92 = 42 + 72 - 2(4)(7)cos R 81 = 16 + 49 -56 cos R 16 = -56 cos R

cos R = -16

56

m R = cos-1 -16

56 107

9

74

T

S

R

Law of Cosines


Simplify

Subtract 65 from both sides.

Solve for cos R

Use the inverse Cosine function to find m/R.

CONFIDENTIAL 17

Now you try!3) Find each measure. Round lengths to the nearest tenth

and angle measures to the nearest degree.

10.5

9.6

5.934

4

10

8

15

10

21

16

18

R

QP

Z

Y

X

L

K

JF

D

E

d) m Rc) YZ

b) m Ka ) DE

3a) 23b) 1

CONFIDENTIAL 18

Engineering Application

The Leaning Tower of Pisa is 56 m tall. In 1999, the tower made a 100˚ angle with the ground. To stabilize the tower, an engineer considered attaching a cable from the top of the tower to a point

that is 40 m from the base. How long would the cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.

40 m

100

56 m

BA

C

Law of Cosines


Find the square root of both sides.

Step 1 Find the length of the cable. AC2 = AB2 + BC2 - 2(AB)(BC)cos B

= 402 + 562 - 2(40)(56)cos 100

AC2 5513.9438

AC 74.3 m

Simplify.

Next page

CONFIDENTIAL 19

Step 2 Find the measure of the angle the cable the cable would make.

sin A

BC =

sin B

AC

sin A

56

sin 100

74.2559

sin A 56 sin 100

74.2559

mA sin-156sin 100

74.2559 48

Law of Sines

Substitute the Calculated values.


Use the inverse sine function to find m/A.

40 m

100

56 m

BA

C

CONFIDENTIAL 20

Now you try!

4) Another engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it

make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.

CONFIDENTIAL 21

Now some practice problems for you!

CONFIDENTIAL 22

Use a calculator to find each trigonometric ratio.

Round to the nearest hundredth.

1) sin 100˚ 2) cos 167˚ 3) tan 92˚

4) tan 141˚ 5) sin 150˚ 6) cos 133˚

CONFIDENTIAL 23


15

70

36

S

T

Ra) RT7)

CONFIDENTIAL 24


10

67

R Q

P

a) m Q

11

131

8

CB

A

b) AB8) 9)

CONFIDENTIAL 25

10) A carpenter makes a triangular frame by joining three pieces of wood that are 20 cm, 24 cm, and 30 cm

long. What are the measures of the angles of the triangle? Round to the nearest degree.

CONFIDENTIAL 26

Finding Trigonometric Ratios for Obtuse Angles

Use a calculator to find each trigonometric ratio. Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.Round to the nearest hundredth.

A sin 135˚ B tan 98 ˚ C cos 108 ˚A sin 135˚ B tan 98 ˚ C cos 108 ˚

sin 135˚ = 0.71sin 135˚ = 0.71 tan 98tan 98˚= -0.72 cos cos 108˚=-0.31

Let’s review

CONFIDENTIAL 27

You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.

In ∆ABC, let h represent the length of the altitude from C to AB.

From the diagram, sin A = (h/b), and sin B = (h/a).

By solving for h, you find that h = bsin A

And h = asin B .So b sin A = asin B,

And (sin A/a) = (sin B/b).

You can use another altitude to show that these ratios

Equal (sin C/c).

‾

h

c

b a

C

BA

CONFIDENTIAL 28

The Law of Sines

Theorem 1.1

For any ABC with side lengths a, b, and c.

sin A

a =

sin B

b =

sin C

c.

ab

cA B

C

CONFIDENTIAL 29

You can use the Law of Sines to solve a triangle if you are given

• two angle measures and any side length (ASA or AAS) or

• two side length and a non-include angle measure (SSA).

CONFIDENTIAL 30

Use the Law of Sines


32

105

F

D

E

Law of Sines

A) DF

sin D

EF =

sin E

DF

sin 105

18 =

sin 32

DF DFsin 105 18sin32

DF = 18sin 32

sin 105 9.9


Cross Products Property

Divide both sides by sin 105˚.

Next page

CONFIDENTIAL 31

B) mS

sin T

RS =

sin S

RT

sin 75

7 =

sin S

5

sin S = 5 sin 75

7

mS sin-15 sin 75

7 44

75

5

7

S

T

R

Law of Sines



Use the inverse sine function to find m/s.

CONFIDENTIAL 32

The Law of Sine cannot be used to solve every triangle. If you know two side lengths and the include angle measure or if you know all three side lengths. You cannot use the Law of Sines. Instead, you can apply the Law of Cosines.

CONFIDENTIAL 33

Theorem 1.2

The Law of Cosines

C

BAc

b a

For any ABC with side lengths a, b, and c: a2 = b2 + c2 - 2bc cosA b2 = a2 + c2 - 2ac cosB

c2 = a2 + b2 - 2ab cosC

CONFIDENTIAL 34

You can use the Law of Cosines to solve a triangle if you are given

• two side lengths and the included angle measure (SAS) or

• three side lengths (SSS).

CONFIDENTIAL 35

Using the Law of Cosines


9

14

62

C

B

A

Law of Cosines


Simplify.

A ) BC

BC2 = AB2 + AC2 - 2(AB)(AC)cos A = 142 + 92 - 2(14)(9)cos 62

BC2 158.6932

BC 12.6 Find the square root of both sides.

Next page

CONFIDENTIAL 36

B ) mR

ST2 = RS2 + RT2 - 2(RS)(RT)cos R

92 = 42 + 72 - 2(4)(7)cos R 81 = 16 + 49 -56 cos R 16 = -56 cos R

cos R = -16

56

m R = cos-1 -16

56 107

9

74

T

S

R

Law of Cosines


Simplify

Subtract 65 from both sides.

Solve for cos R

Use the inverse Cosine function to find m/R.

CONFIDENTIAL 37

Engineering Application

The Leaning Tower of Pisa is 56 m tall. In 1999, the tower made a 100˚ angle with the ground. To stabilize the tower, an engineer considered attaching a cable from the top of the tower to a point that is 40 m from the base. How long would the cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.

40 m

100

56 m

BA

C

Law of Cosines


Find the square root of both sides.

Step 1 Find the length of the cable. AC2 = AB2 + BC2 - 2(AB)(BC)cos B

= 402 + 562 - 2(40)(56)cos 100

AC2 5513.9438

AC 74.3 m

Simplify.

Next page

CONFIDENTIAL 38

Step 2 Find the measure of the angle the cable the cable would make.

sin A

BC =

sin B

AC

sin A

56

sin 100

74.2559

sin A 56 sin 100

74.2559

mA sin-156sin 100

74.2559 48

Law of Sines

Substitute the Calculated values.


Use the inverse sine function to find m/A.

40 m

100

56 m

BA

C

CONFIDENTIAL 39

You did a great job today!You did a great job today!

Geometry Law of Sines and Law of Cosines

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