Transcript bzpc5e_06_01

Chapter 6
Additional Topics
in Trigonometry
6.1 The Law of Sines
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Objectives:
• Use the Law of Sines to solve oblique triangles.
• Use the Law of Sines to solve, if possible, the
triangle or triangles in the ambiguous case.
• Find the area of an oblique triangle using the sine
function.
• Solve applied problems using the Law of Sines.
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Oblique Triangles
An oblique triangle is a triangle that does not contain
a right angle.
An oblique triangle has
either three acute angles
or two acute angles and
one obtuse angle.
The relationships among the sides
and angles of right triangles
defined by the trigonometric
functions are not valid for
oblique triangles.
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The Law of Sines
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Solving Oblique Triangles
Solving an oblique triangle means finding the lengths
of its sides and the measurements of its angles.
The Law of Sines can be used to solve a triangle in
which one side and two angles are known. The three
known measurements can be abbreviated using SAA (a
side and two angles are known) or ASA (two angles and
the side between them are known).
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Example: Solving an SAA Triangle Using the Law of
Sines
Solve the triangle with A = 64°, C = 82°, and c = 14
centimeters. Round lengths of sides to the nearest tenth.
A  B  C  180
64  B  82  180
146  B  180
B  34
a
c

sin A sin C
a sin C  c sin A
c sin A 14sin 64
 12.7 cm
a

sin C
sin82
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Example: Solving an SAA Triangle Using the Law of
Sines (continued)
Solve the triangle with A = 64°, C = 82°, and c = 14
centimeters. Round lengths of sides to the nearest tenth.
B = 34°
a » 12.7 cm
b » 7.9 cm
b
c

sin B sin C
c sin B 14sin 34
» 7.9 cm
b

sin C
sin82
b sin C  c sin B
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Example: Solving an ASA Triangle Using the Law of
Sines
Solve triangle ABC if A = 40°, C = 22.5°, and b = 12.
Round measures to the nearest tenth.
B
A  B  C  180
40  B  22.5  180
62.5  B  180
C
A
12
B  117.5
b
a

b sin A
12sin 40
sin B sin A
 8.7
a

sin B
sin117.5
b sin A  a sin B
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Example: Solving an ASA Triangle Using the Law of
Sines (continued)
Solve triangle ABC if A = 40°, C = 22.5°, and b = 12.
Round measures to the nearest tenth.
B
B  117.5
a  8.7
C
A
c  5.2
12
b
c

sin B sin C
c sin B  b sin C
b sin C 12sin 22.5
 5.2
c

sin B
sin117.5
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The Ambiguous Case (SSA)
If we are given two sides and an angle opposite one of
the two sides (SSA), the given information may result in
one triangle, two triangles, or no triangle at all.
SSA is known as the ambiguous case when using the
Law of Sines because the given information may result
in one triangle, two triangles, or no triangle at all.
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The Ambiguous Case (SSA)
(continued)
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Example: Solving an SSA Triangle Using the Law of Sines
(No Solution)
Solve triangle ABC if A = 50°, a = 10, and b = 20.
There is no angle B for which the sine is greater than 1.
There is no triangle with the given measurements.
a
b

sin A sin B
a sin B  b sin A
b sin A 20sin 50  1.53
sin B 

a
10
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Example: Solving an SSA Triangle Using the Law of Sines
(Two Solutions)
Solve triangle ABC if A = 35°, a = 12, and b = 16. Round
lengths of sides to the nearest tenth and angle measures to
the nearest degree.
a
b

sin A sin B
a sin B  b sin A
b sin A 16sin 35
 0.7648
sin B 

a
12
sin 1 0.7648  50
There are two angles between 0° and
180° for which sinB = 0.7648
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Example: Solving an SSA Triangle Using the Law of Sines
(Two Solutions) (continued)
Solve triangle ABC if A = 35°, a = 12, and b = 16. Round
lengths of sides to the nearest tenth and angle measures to
the nearest degree.
sin 1 0.7648  50
B1  50
A  B1  35  50  85
B2  180  50  130
A  B2  35  130  165
A  B1  180 and A  B2  180,
there are two possible solutions.
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Example: Solving an SSA Triangle Using the Law of Sines
(Two Solutions) (continued)
Solve triangle ABC if A = 35°, a = 12, and b = 16. Round
lengths of sides to the nearest tenth and angle measures to
the nearest degree.
B1  50
B2  130
A  B1  C1  180
35  50  C1  180
85  C1  180
A  B2  C2  180
35  130  C2  180
165  C2  180
C2  15
C1  95
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Example: Solving an SSA Triangle Using the Law of Sines
(Two Solutions) (continued)
Solve triangle ABC if A = 35°, a = 12, and b = 16. Round
lengths of sides to the nearest tenth and angle measures to
the nearest degree.
B2  130 C2  15
B1  50 C1  95
b
c2

b
c1
sin B2 sin C2

sin B1 sin C1
b sin C2  c2 sin B2
b sin C1  c1 sin B1
b sin C2 16sin15
c2 

 5.4
b sin C1 16sin 95
sin B2
sin130
c1 

 20.8
sin B1
sin 50
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Example: Solving an SSA Triangle Using the Law of Sines
(Two Solutions) (continued)
Solve triangle ABC if A = 35°, a = 12, and b = 16. Round
lengths of sides to the nearest tenth and angle measures to
the nearest degree.
There are two triangles. In one triangle, the solution is
B1  50 C1  95 c1  20.8
In the other triangle, the solution is
B2  130 C2  15 c2  5.4
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The Area of an Oblique Triangle
The area of a triangle equals one-half the product of the
lengths of two sides times the sine of their included angle.
In the figure, this wording can be expressed by the
formulas
1
1
1
Area  bc sin A  ab sin C  ac sin B
2
2
2
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Example: Finding the Area of an Oblique Triangle
Find the area of a triangle having two sides of length 8
meters and 12 meters and an included angle of 135°.
Round to the nearest square meter.
C
1
Area  ab sin C
8m
12 m
2
135
A
B
1
 (12)(8)(sin135)
2
 34 sq m
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Example: Application
Two fire-lookout stations are 13 miles apart, with station B
directly east of station A. Both stations spot a fire. The
bearing of the fire from station A is N35°E and the
bearing of the fire from station B is N49°W. How far, to
the nearest tenth of a mile, is the fire from station B?
A  90  35  55
C
B  90  49  41
35
49
A  B  C  180
55  41  C  180
A
B
96  C  180
13
C  84
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Example: Application
(continued)
Two fire-lookout stations are 13 miles apart, with station B directly
east of station A. Both stations spot a fire. The bearing of the fire
from station A is N35°E and the bearing of the fire from station B is
N49°W. How far, to the nearest tenth of a mile, is the fire from
station B?
c
a
sin C
C
35
A

sin A
c sin A  a sin C
c sin A 13sin 55
» 10.7
a

sin C
sin84
49
B
The fire is approximately 10.7 miles from station B.
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