Additional Topics in Trigonometryteachers.dadeschools.net/rcarrasco/bzpc5e_06_01.pdf · Oblique...
Transcript of Additional Topics in Trigonometryteachers.dadeschools.net/rcarrasco/bzpc5e_06_01.pdf · Oblique...
Chapter 6
Additional Topics
in Trigonometry
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1
6.1 The Law of Sines
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 3
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4
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.
180A B C
64 82 180B
146 180B
34B
sin sin
a c
A C
sin sina C c A
sin
sin
c Aa
C
14sin64
sin82
12.7 cm
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7
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.
sin sin
b c
B C
sin sinb C c B
sin
sin
c Bb
C
14sin34
sin82
7.4 cm
34
12.7 cm
7.4 cm
B
a
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.
A
B
C12
180A B C
40 22.5 180B
62.5 180B
117.5B
sin sin
b a
B A
sin sinb A a B
sin
sin
b Aa
B
12sin 40
sin117.5
8.7
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9
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.
A
B
C12
sin sin
b c
B C
sin sinc B b C
sin
sin
b Cc
B
12sin 22.5
sin117.5
5.2
117.5
8.7
5.2
B
a
c
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11
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.
sin sin
a b
A B
sin sina B b A
sinsin
b AB
a
20sin50
10
1.53
There is no angle B for which the sine is greater than 1.
There is no triangle with the given measurements.
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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.
sin sin
a b
A B
sin sina B b A
sinsin
b AB
a
16sin35
12
0.7648
1sin 0.7648 50 There are two angles between 0° and
180° for which sinB = 0.7648
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 14
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.
1sin 0.7648 50
1 50B
2 180 50 130B
1 35 50 85A B
2 35 130 165A B
there are two possible solutions.
1 2180 and 180 ,A B A B
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 15
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.
1 50B 2 130B
1 1 180A B C 2 2 180A B C
135 50 180C
185 180C
1 95C
235 130 180C
2165 180C
2 15C
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 16
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.
1 50B 1 95C 2 130B 2 15C
1
1 1sin sin
b c
B C
2
2 2sin sin
b c
B C
1 1 1sin sinb C c B2 2 2sin sinb C c B
11
1
sin
sin
b Cc
B
22
2
sin
sin
b Cc
B
16sin95
sin50
20.8
16sin15
sin130
5.4
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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
In the other triangle, the solution is
1 50B 1 95C 1 20.8c
2 130B 2 15C 2 5.4c
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 18
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 1Area sin sin sin
2 2 2bc A ab C ac B
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 19
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.
8 m 12 m135
A B
C1
Area sin2
ab C
1(12)(8)(sin135 )
2
34 sq m
Copyright © 2014, 2010, 2007 Pearson Education, Inc. 20
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 B
C
3549
13
90 35 55A
90 49 41B
180A B C
55 41 180C
96 180C 84C
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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?
A B
C
3549
sin sin
c a
C A
sin sinc A a C
sin
sin
c Aa
C
13sin55
sin84
11
The fire is approximately 11 miles from station B.