Law of Sines

Law of Sines

Objective: To solve triangles that are not right triangles

Law of Sines

• We have been solving for sides and angles of right triangles with a variety of methods. We will now look at solving triangles that are not right triangles. These triangles are called oblique triangles.

Law of Sines

• We have been solving for sides and angles of right triangles using a variety of methods. We will now look at solving triangles that are not right triangles. These triangles are called oblique triangles.

• To solve an oblique triangle with the Law of Sines, you need to know the measure of at least one side and the opposite angle. This breaks down into the following cases.

Law of Sines

• To use the Law of Sines, you need to have:1. Two angles and any side (AAS or ASA)2. Two sides and an angle opposite one of them (SSA)

Law of Sines

• Given triangle ABC with sides a, b and c, then:

c

C

b

B

a

A sinsinsin

Example 1(AAS)

• For the given triangle, find the remaining angle and sides.

Example 1(AAS)


• We know that <B=28.70

• We know that <C=102.30

• We can find that <A=490

Example 1(AAS)


ca

ooo 3.102sin

4.27

7.28sin49sin

Example 1(AAS)


ca

ooo 3.102sin

4.27

7.28sin49sin

017.49sin

a

o

ao

017.

49sin

aft 39.44

Example 1(AAS)


ca

ooo 3.102sin

4.27

7.28sin49sin

017.3.102sin

c

o

017.49sin

a

o

ao

017.

49sinc

o

017.

3.102sin

aft 39.44 cft 47.57

You Try

• Given triangle ABC, find the missing angle and sides.

You Try

• Given triangle ABC, find the missing angle and sides.• <A = 350

ba

ooo 40sin35sin

20

105sin

a

o35sin048.

048.

35sin o

a

95.11a

You Try


ba

ooo 40sin35sin

20

105sin

a

o35sin048.

048.

35sin o

a

95.11a

b

o40sin048.

048.

40sin o

b

39.13b

Example 2(ASA)

• A pole tilts toward the sun at an 80 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 430. How tall is the pole?

Example 2(ASA)


• We know that <A=430

• We know that <B=980

• We can find that <C=390

Example 2(ASA)


22

39sin98sin43sin ooo

ba

Example 2(ASA)


029.98sin

b

o22

39sin98sin43sin ooo

ba

bo

029.

98sin

bft 15.34

Example 2(ASA)


029.98sin

b

o22

39sin98sin43sin ooo

ba

bo

029.

98sin

bft 15.34

029.43sin

a

o

ao

029.

43sin

aft 52.23

You Try


• <B = 440

• c = 12 ft

You Try


• <B = 440

• c = 12 ft• <C = 660

a

o70sin076.

ba

ooo 44sin70sin

12

66sin

076.

70sin o

a

fta 36.12

You Try


• <B = 440

• c = 12 ft• <C = 660

a

o70sin076.

ba

ooo 44sin70sin

12

66sin

076.

70sin o

a

fta 36.12

b

o44sin076.

076.

44sin o

b

ftb 14.9

Example 3

• We know from Geometry that SSA does not make a unique triangle. When given SSA, one of three situations may occur.

1. One unique triangle2. No triangle3. Two different triangles

Example 3

• For the given triangle, find the missing angles and side.

Example 3


Bo

sin22

42sin12

c

CBo sin

12

sin

22

42sin

Bsin365.

B 365.sin 1

Bo 40.21

Example 3


Bo

sin22

42sin12

c

CBo sin

12

sin

22

42sin

Bsin365.

B 365.sin 1

Bo 40.21

c

oo 6.116sin

22

42sin

030.

6.116sin o

c

inc 80.29

Example 3

• Remember, there are two answers to• <B = 21.40 or <B = 158.60.• The answer of 158.60 won’t work since this angle

added to the given angle of 420 would be greater than 1800, and we know that doesn’t make sense for a triangle, therefore there is only one solution to this problem.

B 365.sin 1

Example 4


Example 4


• There is no solution to this.

Bo

sin15

85sin25

c

CBo sin

25

sin

15

85sin

Bsin66.1

B 66.1sin 1

Example 5


• a = 12m• b = 31m• <A = 20.50

|||||||

Example 5


• A = 12m • B = 31m• <A = 20.50

Bo

sin12

5.20sin31

c

CBo sin

31

sin

12

5.20sin

Bsin905.

B 905.sin 1

Bo 8.64

Example 5


• A = 12m • B = 31m• <A = 20.50

Bo

sin12

5.20sin31

c

CBo sin

31

sin

12

5.20sin

Bsin905.

B 905.sin 1

c

oo 7.94sin

12

5.20sin

029.

7.94sin o

c

mc 37.34Bo 8.64

Example 5

• Remember, there are two answers to .• <B = 64.80 or 115.20.• Since both answers work with an angle of 20.50, there

are two triangles possible for this problem.

B 905.sin 1

Example 5


• A = 12m • B = 31m• <A = 20.50

Bo

sin12

5.20sin31

c

CBo sin

31

sin

12

5.20sin

Bsin905.

B 905.sin 1

Bo 2.115

Example 5


• A = 12m • B = 31m• <A = 20.50

Bo

sin12

5.20sin31

c

CBo sin

31

sin

12

5.20sin

Bsin905.

B 905.sin 1

c

oo 3.44sin

12

5.20sin

029.

3.44sin o

c

mc 08.24Bo 2.115

Example 5

• Here are the two triangles together.

You Try

• Page 598

• 20

You Try

• Page 598

• 20

There is no triangle

Bo

sin125

110sin200

c

CBo sin

200

sin

125

110sin

Bsin50.1

B 50.1sin 1

You Try

• Page 598

• 22

You Try

• Page 598

• 22

Bo

sin34

76sin21

c

CBo sin

21

sin

34

76sin

Bsin599.

B 599.sin 1

B8.36

You Try

• Page 598

• 22

Bo

sin34

76sin21

c

CBo sin

21

sin

34

76sin

Bsin599.

B 599.sin 1

B8.36

c

oo 2.67sin

34

76sin

028.

2.67sin o

c

92.32c

You Try

• Page 598• 22• If <B = 36.80, it can also be 143.20. This with the given

angle of 760 is more than 1800, so there is only one triangle.

You Try

• Page 598• 6

You Try

• Page 598• 6

10

sinsin

9

60sin C

b

Bo

Co

sin9

60sin10

2.74

962.sin 1

C

C

b

o 8.45sin

9

60sin

096.

8.45sinb

47.7b

Example 5

• Remember, there are two answers to .• <B = 74.20 or 105.80.• Since both answers work with an angle of 600, there

are two triangles possible for this problem.

B 962.sin 1

You Try

• Page 598• 6

10

sinsin

9

60sin C

b

Bo

Co

sin9

60sin10

8.105

962.sin 1

C

C

b

o 2.14sin

9

60sin

096.

2.14sinb

55.2b

Area of an Oblique Triangle

• In the past, we needed the height of a triangle in order to find the area. We will now use an equation to find area that doesn’t need the height. There are three forms of this equation. They are:



BacAbcCabarea sin2

1sin

2

1sin

2

1



• In words, this equation says that area is equal to: ½(side)(side)(sin of the included angle)

BacAbcCabarea sin2

1sin

2

1sin

2

1

Example 6

• Find the area of a triangular lot having two sides of length 90 meters and 52 meters and an included angle of 1020.

Example 6

• Find the area of a triangular lot having two sides of length 90 meters and 52 meters and an included angle of 1020.

22289marea

)102)(sin52)(90(2

1 oarea

Homework

• Page 598

• 1-7 odd

• 13, 17,19, 21, 23

• 29, 30

Law of Sines

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