Download Law of Sines - Souderton Math

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:
sin A sin B sin C


a
b
c
Example 1(AAS)
• For the given triangle, find the remaining angle and
sides.
Example 1(AAS)
• For the given triangle, find the remaining angle and
sides.
• We know that <B=28.70
• We know that <C=102.30
• We can find that <A=490
Example 1(AAS)
• For the given triangle, find the remaining angle and
sides.
sin 49o sin 28.7 o sin 102.3o


a
27.4
c
Example 1(AAS)
• For the given triangle, find the remaining angle and
sides.
sin 49o sin 28.7 o sin 102.3o


a
27.4
c
sin 49o
 .017
a
sin 49o
a
.017
44.39 ft  a
Example 1(AAS)
• For the given triangle, find the remaining angle and
sides.
sin 49o sin 28.7 o sin 102.3o


a
27.4
c
sin 49o
 .017
a
sin 49o
a
.017
44.39 ft  a
sin 102.3o
 .017
c
sin 102.3o
c
.017
57.47 ft  c
You Try
• Given triangle ABC, find the missing angle and sides.
You Try
• Given triangle ABC, find the missing angle and sides.
• <A = 350
sin 105o sin 35o sin 40o


20
a
b
sin 35o
.048 
a
sin 35o
a
.048
a  11.95
You Try
• Given triangle ABC, find the missing angle and sides.
• <A = 350
sin 105o sin 35o sin 40o


20
a
b
sin 35o
.048 
a
sin 40o
.048 
b
sin 35o
a
.048
sin 40o
b
.048
a  11.95
b  13.39
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)
• 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?
• We know that <A=430
• We know that <B=980
• We can find that <C=390
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?
sin 43o sin 98o sin 39o


a
b
22
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?
sin 98o
 .029
b
sin 98o
b
.029
34.15 ft  b
sin 43o sin 98o sin 39o


a
b
22
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?
sin 98o
 .029
b
sin 98o
b
.029
34.15 ft  b
sin 43o sin 98o sin 39o


a
b
22
sin 43o
 .029
a
sin 43o
a
.029
23.52 ft  a
You Try
•
•
•
•
Given triangle ABC, find the missing angle and sides.
<A = 700
<B = 440
c = 12 ft
You Try
•
•
•
•
•
Given triangle ABC, find the missing angle and sides.
<A = 700
<B = 440
c = 12 ft
<C = 660
sin 66o sin 70o sin 44o


12
a
b
.076 
sin 70
a
o
sin 70o
a
.076
a  12.36 ft
You Try
•
•
•
•
•
Given triangle ABC, find the missing angle and sides.
<A = 700
<B = 440
c = 12 ft
<C = 660
sin 66o sin 70o sin 44o


12
a
b
.076 
sin 70
a
o
sin 70o
a
.076
a  12.36 ft
sin 44o
.076 
b
sin 44o
b
.076
b  9.14 ft
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 triangle
2. No triangle
3. Two different triangles
Example 3
• For the given triangle, find the missing angles and
side.
Example 3
• For the given triangle, find the missing angles and
side.
sin 42o sin B sin C


22
12
c
12 sin 42o
 sin B
22
.365  sin B
sin 1 .365  B
21.40 o  B
Example 3
• For the given triangle, find the missing angles and
side.
sin 42o sin B sin C


22
12
c
12 sin 42o
 sin B
22
.365  sin B
sin 1 .365  B
21.40 o  B
sin 42o sin 116.6o

22
c
sin 116.6o
c
.030
c  29.80in
Example 3
• Remember, there are two answers to sin 1 .365  B
• <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.
Example 4
• For the given triangle, find the missing angles and
side.
Example 4
• For the given triangle, find the missing angles and
side.
sin 85o sin B sin C


15
25
c
25 sin 85o
 sin B
15
1.66  sin B
sin 1 1.66  B
• There is no solution to this.
Example 5
• For the given triangle, find the missing angles and
side.
• a = 12m
|||||||
• b = 31m
• <A = 20.50
Example 5
• For the given triangle, find the missing angles and
side.
o
sin
20
.
5
sin B sin C
• A = 12m


12
31
c
• B = 31m
• <A = 20.50 31sin 20.5  sin B
o
12
.905  sin B
sin 1 .905  B
64.8o  B
Example 5
• For the given triangle, find the missing angles and
side.
o
sin
20
.
5
sin B sin C
• A = 12m


12
31
c
• B = 31m
sin 20.5
sin 94.7
• <A = 20.50 31sin 20.5  sin B

o
12
o
12
.905  sin B
o
c
sin 94.7 o
c
.029
sin 1 .905  B
64.8  B
o
c  34.37m
Example 5
• Remember, there are two answers to sin 1 .905  B .
• <B = 64.80 or 115.20.
• Since both answers work with an angle of 20.50, there
are two triangles possible for this problem.
Example 5
• For the given triangle, find the missing angles and
side.
o
sin
20
.
5
sin B sin C
• A = 12m


12
31
c
• B = 31m
• <A = 20.50 31sin 20.5  sin B
o
12
.905  sin B
sin 1 .905  B
115.2o  B
Example 5
• For the given triangle, find the missing angles and
side.
o
sin
20
.
5
sin B sin C
• A = 12m


12
31
c
• B = 31m
• <A = 20.50 31sin 20.5  sin B
sin 20.5o sin 44.3o
o
12
.905  sin B
sin 1 .905  B
115.2o  B
12

c
sin 44.3o
c
.029
c  24.08m
Example 5
• Here are the two triangles together.
You Try
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• 20
You Try
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• 20
sin 110o sin B sin C


125
200
c
200 sin 110o
 sin B
125
1.50  sin B
sin 1 1.50  B
There is no triangle
You Try
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• 22
You Try
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• 22
sin 76o sin B sin C


34
21
c
21sin 76o
 sin B
34
.599  sin B
sin 1 .599  B
36.8  B
You Try
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• 22
sin 76o sin B sin C


34
21
c
21sin 76o
 sin B
34
.599  sin B
sin 1 .599  B
36.8  B
sin 76o sin 67.2o

34
c
sin 67.2o
c
.028
c  32.92
You Try
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• 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
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• 6
You Try
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• 6
o
sin 60
sin B sin C


9
b
10
10 sin 60o
 sin C
9
sin 1 .962  C
 C  74.2
sin 60o sin 45.8

9
b
b
sin 45.8
.096
b  7.47
Example 5
• Remember, there are two answers to sin 1 .962  B .
• <B = 74.20 or 105.80.
• Since both answers work with an angle of 600, there
are two triangles possible for this problem.
You Try
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• 6
o
sin 60
sin B sin C


9
b
10
10 sin 60o
 sin C
9
sin 1 .962  C
 C  105.8
sin 60o sin 14.2

9
b
b
sin 14.2
.096
b  2.55
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:
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:
1
1
1
area  ab sin C  bc sin A  ac sin B
2
2
2
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:
1
1
1
area  ab sin C  bc sin A  ac sin B
2
2
2
• In words, this equation says that area is equal to:
½(side)(side)(sin of the included angle)
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.
area 
1
(90)(52)(sin 102o )
2
area  2289m 2
Homework
•
•
•
•
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1-7 odd
13, 17,19, 21, 23
29, 30