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Copyright © 2011 Pearson Education, Inc.
Slide 10.1-1
Chapter 10: Applications of Trigonometry
and Vectors
10.1 The Law of Sines
10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex
Numbers
10.5 Powers and Roots of Complex Numbers
10.6 Polar Equations and Graphs
10.7 More Parametric Equations
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-2
10.1 The Law of Sines
Congruence Axioms
Side-Angle-Side (SAS)
If two sides and the included angle of one
triangle are equal, respectively, to two sides and
the included angle of a second triangle, then the
triangles are congruent.
Angle-Side-Angle (ASA) If two angles and the included side of one
triangle are equal, respectively, to two angles and
the included side of a second triangle, then the
triangles are congruent.
Side-Side-Side (SSS)
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If three sides of one triangle are equal to three
sides of a second triangle, the triangles are
congruent.
Slide 10.1-3
10.1 Data Required for Solving
Oblique Triangles
Case 1
Case 2
Case 3
Case 4
One side and two angles are known
(SAA or ASA).
Two sides and one angle not included
between the two sides are known (SSA).
This case may lead to zero, one, or two
triangles.
Two sides and the angle included between
the two sides are known
(SAS).
Three sides are known
(SSS).
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-4
10.1 Derivation of the Law of Sines
Start with an acute or obtuse triangle
and construct the perpendicular from
B to side AC. Let h be the height of this
perpendicular. Then c and a are the
hypotenuses of right triangle ADB
and BDC, respectively.
h
sin A 
or h  c sin A,
c
h
sin C 
or h  a sin C
a
a
c
a sin C  c sin A 

sin A sin C
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-5
10.1 The Law of Sines
In a similar way, by constructing perpendiculars from
other vertices, the following theorem can be proven.
Law of Sines
In any triangle ABC, with sides a, b, and c,
a
b
c


.
sin A sin B sin C
Alternative forms are sometimes convenient to use:
sin A sin B sin C


.
a
b
c
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-6
10.1 Using the Law of Sines to Solve a
Triangle
Example Solve triangle ABC if A = 32.0°, B = 81.8°,
and a = 42.9 centimeters.
Solution Draw the triangle
and label the known values.
Because A, B, and a are known,
we can apply the law of sines involving these variables.
a
b

sin A sin B
42.9
b

 b  80.1 cm


sin 32.0 sin 81.8
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-7
10.1 Using the Law of Sines to Solve a
Triangle
To find C, use the fact that there are 180° in a triangle.
A  B  C  180
C  180  A  C
 180  32.0  81.8  66.2

Now we can find c.
a
c

sin A sin C
42.9
c



sin 32.0 sin 66.2
Copyright © 2011 Pearson Education, Inc.



 c  74.1 cm
Slide 10.1-8
10.1 Using the Law of Sines in an
Application (ASA)
Example Two stations are on an east-west
line 110 miles apart. A forest fire is located
on a bearing of N 42° E from the western
station at A and a bearing of N 15° E from
the eastern station at B. How far is the fire
from the western station?
Solution
Angle BAC = 90° – 42° = 48°
Angle B = 90° + 15° = 105°
Angle C = 180° – 105° – 48° = 27°
Using the law of sines to find b gives
b
110


sin 105 sin 27 
b  234 miles.
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-9
10.1 Ambiguous Case
Applying the Law of Sines
1. For any angle , –1  sin   1, if sin  = 1, then
 = 90° and the triangle is a right triangle.
2. sin  = sin(180° –  ). (Supplementary angles
have the same sine value.)
3. The smallest angle is opposite the shortest side, the
largest angle is opposite the longest side, and the
middle-value angle is opposite the intermediate
side (assuming that the triangle has sides that are
all of different lengths).
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Slide 10.1-10
10.1 Ambiguous Case
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Slide 10.1-11
10.1 Ambiguous Case for Obtuse Angle A
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Slide 10.1-12
10.1 Solving the Ambiguous Case:
No Such Triangle
Example Solve the triangle ABC if B = 55°40´,
b = 8.94 meters, and a = 25.1 meters.
Solution Use the law of sines to find A.
sin A sin B

a
b
sin A sin 55 40'

25.1
8.94
sin A  2.3184379
Since sin A cannot be greater than 1, the triangle does
not exist.
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-13
10.1 Solving the Ambiguous Case:
Two Triangles
Example
Solve the triangle ABC if A = 55.3°,
a = 22.8 feet, and b = 24.9 feet.
Solution
sin A sin B

a
b
sin 55.3 sin B

22.8
24.9
sin B  .8978678  B1  63.9
B2  180  63.9  116.1
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-14
10.1 Solving the Ambiguous Case:
Two Triangles
To see if B2 = 116.1° is a valid possibility, add 116.1°
to the measure of A: 116.1° + 55.3° = 171.4°. Since
this sum is less than 180°, it is a valid triangle.
Now separate the triangles into two: AB1C1 and AB2C2.
C1  180  A  B1  180  55.3  63.9  60.8





a
c1

sin A sin C1
22.8
c1


sin 55.3 sin 60.8
c1  24.2 feet
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-15
10.1 Solving the Ambiguous Case:
Two Triangles
Now solve for triangle AB2C2.
C2  180  A  B2  180  55.3  116.1  8.6
a
c2

sin A sin C2
22.8
c2


sin 55.3 sin 8.6
c2  4.15 feet
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-16
10.1 Number of Triangles Satisfying the
Ambiguous Case (SSA)
Let sides a and b and angle A be given in triangle ABC. (The
law of sines can be used to calculate sin B.)
1. If sin B > 1, then no triangle satisfies the given conditions.
2. If sin B = 1, then one triangle satisfies the given conditions
and B = 90°.
3. If 0 < sin B < 1, then either one or two triangles satisfy the
given conditions
(a) If sin B = k, then let B1 = sin-1 k and use B1 for B in the
first triangle.
(b) Let B2 = 180° – B1. If A + B2 < 180°, then a second
triangle exists. In this case, use B2 for B in the second
triangle.
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-17
10.1 Solving the Ambiguous Case:
One Triangle
Example Solve the triangle ABC, given A = 43.5°,
a = 10.7 inches, and c = 7.2 inches.
Solution
sin C sin 43.5

7.2
10.7
sin C  .46319186  C  27.6
The other possible value for C:
C = 180° – 27.6° = 152.4°.
Add this to A: 152.4° + 43.5° = 195.9° > 180°
Therefore, there can be only one triangle.
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-18
10.1 Solving the Ambiguous Case:
One Triangle
B  180  27.6  43.5  108.9
b
10.7


sin 108.9 sin 43.5
b  14.7 inches
Copyright © 2011 Pearson Education, Inc.
Slide 10.1-19