The Law of SinesSection 6.1

Mr. Thompson

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An oblique triangle is a triangle that has no right angles.

To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side.

C

BA

ab

c

3

The following cases are considered when solving oblique triangles.

1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

A

C

c

A

B

c

a

cb

C

c

a

c

aB

The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.)

Law of Sines

If ABC is an oblique triangle with sides a, b, and c, then

.sin sin sin

a b cA B C

Acute Triangle

C

BA

bh

c

a

C

BA

bh

c

a

Obtuse Triangle

Find the remaining angle and sides of the triangle.Example (ASA):

sin sina b

A B

The third angle in the triangle is A = 180 – A – B = 180 – 10 – 60 = 110.

C

BA

b

c

60

10

a = 4.5 ft

110

4.5110 60sin sin

b

Use the Law of Sines to find side b and c.

4.15 feetb

4.15 ft

sin sina c

A C

4.5110 10sin sin

c

0.83 feetc

0.83 ft

For the triangle in the figure, C = 102.3°, B = 28.7°, and b = 27.4 feet. Find the remaining angle and sides.

Ex. 2. A pole tilts toward the sun at an 8° angle from the vertical, and it casts a 22-foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43°. How tall is the pole?

Remember back to Geometry-Which cases were enough to prove two triangles congruent?-Why was there a problem with Some of the cases?

Try to come up with counterexamples for these cases

The Ambiguous Case (SSA) TableKnowing two sides and an angle opposite one of these

sides might not be enough to form a triangle. This information may allow you to form 0, 1, or 2 triangles!

Use the Law of Sines to solve the triangle.A = 110, a = 125 inches, b = 100 inches

Example (SSA):

sin sina b

A B

125 100110sin sin B

48.74B

C 180 – 110 – 48.74 sin sin

a cA C

125110 21.26sin sin

c 48.23 inchesc

C

BA

b = 100 in

c

a = 125 in

110 48.74

21.26

48.23 in

= 21.26

Use the Law of Sines to solve the triangle.A = 76, a = 18 inches, b = 20 inches

Example (SSA):

sin sina b

A B

18 2076sin sin B

sin 1.078B

There is no angle whose sine is 1.078.

There is no triangle satisfying the given conditions.

C

AB

b = 20 ina = 18 in

76

Show that there is no triangle for which a = 15, b = 25, and A = 85°.

Use the Law of Sines to solve the triangle.A = 58, a = 11.4 cm, b = 12.8 cm

Example (SSA):

sin sina b

A B

11.4 12.858sin sin B

72.2

10.3 cm

Two different triangles can be formed.

49.8

a = 11.4 cm

C

AB1

b = 12.8 cm

c

58

Example continues.

1 72.2B

12.si

8n si49.8 2.2n 7

c

10.3c C 180 – 58 – 72.2 = 49.8

sin sinc b

C B

Use the Law of Sines to solve the second triangle.A = 58, a = 11.4 cm, b = 12.8 cm

Example (SSA) continued:

B2 180 – 72.2 = 107.8

107.8

C

AB2

b = 12.8 cm

c

a = 11.4 cm

58

14.2

3.3 cm

72.2

10.3 cm

49.8

a = 11.4 cm

C

AB1

b = 12.8 cm

c

58

C 180 – 58 – 107.8 = 14.2

12.si

8n si14.2 2.2n 7

c

3.3c

sin sinc b

C B

Find two triangles for which a = 12 meters, b = 31 meters, and A = 20.5°.

My Old HouseArt Museum

City Hall

1.25 miles

Area of an Oblique Triangle

C

BA

b

c

aFind the area of the triangle.A = 74, b = 103 inches, c = 58 inches

Example:

74

103 in

58 in 1Area = sin

2bc A

1 1 1Area sin sin sin2 2 2

bc A ab C ac B

103 51= ( )( )8 sin2

74

2871 square inches

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The following cases are considered when solving oblique triangles.

1. Two angles and any side (AAS or ASA)

2. Two sides and an angle opposite one of them (SSA)

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

A

C

c

A

B

c

a

cb

C

c

a

c

aB

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The last two cases (SSS and SAS) can be solved using the Law of Cosines. (The first two cases can be solved using the Law of Sines.)

Law of CosinesStandard Form

2 2 2 2 cosa b c bc A

Alternative Form2 2 2

cos2

b c aAbc

2 2 2 2 cosb a c ac B 2 2 2

cos2

a c bBac

2 2 2 2 cosc a b ab C 2 2 2

cos2

a b cCab

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Find the three angles of the triangle.Example:

C

BA

86

12

2 2 2cos

2a b cC

ab

Find the angle opposite the longest side first.

2 2 2122(

8 68 6)( )

64 36 14496

4496

117.3C

117.12

sin 3 in6

s BLaw of Sines: 26.4B

180 117.3 26.4A

36.3

117.3

26.4

36.3

21

Solve the triangle.Example:

67.8

759.9 6.2

sin sin A Law of Sines: 37.2A

180 75 37.2 67.8C

37.2

C

BA

6.2

759.52 2 2 2 cosb a c ac B

2 2( ) ( ) 2( )6.2 9.5 6.2 9.5( 7s 5)co

38.44 90.25 (117.8)(0.25882)

98.209.9b

9.9

Law of Cosines:

Example 1: Find the three angles of the triangle shown.

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Example 2: Find the remaining angles and sides of the triangle shown.

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Heron’s Area Formula

Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by

Area ( )( )( )s s a s b s c

Example:Find the area of the triangle.

11.5 82

52

10a b cs

Area ( 5)(11.5 11.5 11.5 11.8)( 10)5

5

108where .

2a b cs

19.8 square units

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Application:Two ships leave a port at 9 A.M. One travels at a bearing of N 53 W at 12 mph, and the other travels at a bearing of S 67 W at 16 mph. How far apart will the ships be at noon?

53

67

c

36 mi

48 mi

C

At noon, the ships have traveled for 3 hours.

Angle C = 180 – 53 – 67 = 60

2 2 2 2 cosc a b ab C 2 2 2( )( ) c36 4 os 68 36 8 1 24 80 7

43 mic

The ships will be approximately 43 miles apart.

43 mi 60

N

Example 5: Find the area of a triangle having sides a = 43 meters, b = 53 meters, and c = 72 meters.

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