6.1 Law of Sines Objective To use Law of Sines to solve oblique triangles and to find the areas of oblique triangles.
6.1 Law of Sines Objective To use Law of Sines to solve oblique triangles and to find the areas of oblique triangles.
Dec 13, 2015
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6.1 Law of Sines
Objective
To use Law of Sines to solve oblique triangles and to find the areas of
oblique triangles.
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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
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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 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
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Find the remaining angle and sides of the triangle.Example 1 (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
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Use the Law of Sines to solve the triangle.A = 110, a = 125 inches, b = 100 inches
Example 2 (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
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The Ambiguous Case (SSA)
• Two angles and one side determine a unique triangle. But what if you are given two sides and one opposite angle? Three possible situations can occur:
• 1) no such triangle exists;• 2) one such triangle exits; • 3) two distinct triangles may satisfy the
conditions.
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The Ambiguous Case (SSA
A is acutea < b
a > b * sin A
Two Solutions
a = b * sin A
One Solution
a < b * sin A
No Solutions
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The Ambiguous Case (SSA)
• A is acute and a ≥ b
• One Solution
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The Ambiguous Case (SSA)
A is obtuse
a > b
One Solution
a < b
No Solution
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Use the Law of Sines to solve the triangle.A = 76, a = 18 inches, b = 20 inches
Example 3 (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
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Use the Law of Sines to solve the triangle.A = 58, a = 11.4 cm, b = 12.8 cm
Example 4 (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
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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
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Area of an Oblique Triangle
C
BA
b
c
aFind the area of the triangle.A = 74, b = 103 inches, c = 58 inches
Example 5:
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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Example 6 Finding the Area of a Triangular Lot
• Find the area of a triangular lot containing side lengths that measure 24 yards and 18 yards and form an angle of 80°
• A = ½(18)(24)sin80
• A = 212.7 yards
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A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14 with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20. How tall is the flagpole?
Example 7 Application:
20
Flagpole height: b70
34
16 m
14
sin sina b
A B
1670si 34n sin
b
9.5 metersb
The flagpole is approximately 9.5 meters tall.
B
A
C
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