Download 8.1

8
Applications
of
Trigonometry
Copyright © 2009 Pearson Addison-Wesley
8.1-1
8 Applications of Trigonometry
8.1 The Law of Sines
8.2 The Law of Cosines
8.3 Vectors, Operations, and the Dot Product
8.4 Applications of Vectors
8.5 Trigonometric (Polar) Form of Complex
Numbers; Products and Quotients
8.4 De Moivre’s Theorem; Powers and Roots of
Complex Numbers
8.5 Polar Equations and Graphs
8.6 Parametric Equations, Graphs, and
Applications
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8.1 The Law of Sines
Congruency and Oblique Triangles ▪ Derivation of the Law of
Sines ▪ Using the Law of Sines ▪ Ambiguous Case ▪ Area of a
Triangle
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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.
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Congruence Axioms
Side-Side-Side (SSS)
If three sides of one triangle are equal,
respectively, to three sides of a second
triangle, then the triangles are
congruent.
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Oblique Triangles
Oblique triangle
A triangle that is not a right
triangle
The measures of the three sides and the three
angles of a triangle can be found if at least one
side and any other two measures are known.
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Data Required for Solving Oblique Triangles
Case 1 One side and two angles are
known (SAA or ASA).
Case 2 Two sides and one angle not
included between the two sides
are known (SSA). This case may
lead to more than one triangle.
Case 3 Two sides and the angle included
between the two sides are known
(SAS).
Case 4 Three sides are known (SSS).
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Note
If three angles of a triangle are
known, unique side lengths cannot
be found because AAA assures only
similarity, not congruence.
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Derivation of the Law of Sines
Start with an oblique triangle,
either acute or obtuse.
Let h be the length of the
perpendicular from vertex B
to side AC (or its extension).
Then c is the hypotenuse of
right triangle ABD, and a is
the hypotenuse of right
triangle BDC.
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Derivation of the Law of Sines
In triangle ADB,
𝒉
𝐬𝐢𝐧 𝑪 = 𝒐𝒓 𝒉 = 𝒂 𝐬𝐢𝐧 𝑪
𝒂
Since h = c sin A and h = a sin C,
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Derivation of the Law of Sines
In triangle BDC,
Similarly, it can be shown that
and
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Law of Sines
In any triangle ABC, with sides a, b, and c,
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Example 1
USING THE LAW OF SINES TO SOLVE A
TRIANGLE (SAA)
Solve triangle ABC if
A = 32.0°, C = 81.8°,
and a = 42.9 cm.
Law of sines
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Example 1
USING THE LAW OF SINES TO SOLVE A
TRIANGLE (SAA) (continued)
A + B + C = 180°
C = 180° – A – B
C = 180° – 32.0° – 81.8° = 66.2°
Use the Law of Sines to find c.
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Example 2
USING THE LAW OF SINES IN AN
APPLICATION (ASA)
Kurt wishes to measure the distance across the Big
Muddy River. He determines that C = 112.90°,
A = 31.10°, and b = 347.6 ft. Find the distance a
across the river.
First find the measure of angle B.
B = 180° – A – C = 180° – 31.10° – 112.90° = 36.00°
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Example 2
USING THE LAW OF SINES IN AN
APPLICATION (ASA) (continued)
Now use the Law of Sines to find the length of side a.
The distance across the river is about 305.5 feet.
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Example 3
USING THE LAW OF SINES IN AN
APPLICATION (ASA)
Two ranger stations are on an
east-west line 110 mi apart. A
forest fire is located on a bearing
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?
First, find the measures of the angles in the triangle.
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Example 3
USING THE LAW OF SINES IN AN
APPLICATION (ASA) (continued)
Now use the Law of Sines to find b.
The fire is about 234 miles from the western station.
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Description of the Ambiguous
Case
If the lengths of two sides and the angle opposite
one of them are given (Case 2, SSA), then zero,
one, or two such triangles may exist.
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If A is acute, there are four possible outcomes.
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If A is obtuse, there are two possible outcomes.
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Applying the Law of Sines
1. For any angle θ of a triangle,
0 < sin θ ≤ 1. If sin θ = 1, then θ = 90°
and the triangle is a right triangle.
2. sin θ = sin(180° – θ) (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 the triangle
has sides that are all of different
lengths).
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Example 4
SOLVING THE AMBIGUOUS CASE (NO
SUCH TRIANGLE)
Solve triangle ABC if B = 55°40′, b = 8.94 m, and
a = 25.1 m.
Law of sines
(alternative form)
Since sin A > 1 is impossible, no such triangle exists.
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Example 4
SOLVING THE AMBIGUOUS CASE (NO
SUCH TRIANGLE) (continued)
An attempt to sketch the triangle leads to this figure.
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Note
In the ambiguous case, we are given
two sides and an angle opposite one of
the sides (SSA).
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Example 5
SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES)
Solve triangle ABC if A = 55.3°, a = 22.8 ft, and
b = 24.9 ft.
There are two angles between 0° and 180° such that
sin B ≈ .897867:
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Example 5
SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
Solve separately for triangles
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Example 5
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SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
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Example 5
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SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
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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 the value of sin B.)
1. If applying the law of sines results in an
equation having 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°.
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Number of Triangles Satisfying the
Ambiguous Case (SSA)
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.
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SOLVING THE AMBIGUOUS CASE (ONE
TRIANGLE)
Example 6
Solve triangle ABC given A = 43.5°, a = 10.7 in., and
c = 7.2 in.
There is another angle C with sine value .46319186:
C = 180° – 27.6° = 152.4°
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Example 6
SOLVING THE AMBIGUOUS CASE (ONE
TRIANGLE) (continued)
Since c < a, C must be
less than A. So C = 152.4°
is not possible.
B = 180° – 27.6° – 43.5° = 108.9°
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Example 7
ANALYZING DATA INVOLVING AN
OBTUSE ANGLE
Without using the law of sines, explain why A = 104°,
a = 26.8 m, and b = 31.3 m cannot be valid for a
triangle ABC.
Because A is an obtuse angle, it must be the largest
angle of the triangle. Thus, a must be the longest side
of the triangle.
We are given that b > a, so no such triangle exists.
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Area of a Triangle (SAS)
In any triangle ABC, the area A is given by
the following formulas:
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Note
If the included angle measures 90°, its
sine is 1, and the formula becomes the
familiar
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Example 8
FINDING THE AREA OF A TRIANGLE
(SAS)
Find the area of triangle ABC.
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Example 9
FINDING THE AREA OF A TRIANGLE
(ASA)
Find the area of triangle ABC if A = 24°40′,
b = 27.3 cm, and C = 52°40′.
Draw a diagram.
Before the area formula can be used, we must find
either a or c.
B = 180° – 24°40′ – 52°40′ = 102°40′
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Example 9
FINDING THE AREA OF A TRIANGLE
(ASA) (continued)
Now find the area.
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Caution
Whenever possible, use given values in
solving triangles or finding areas rather
than values obtained in intermediate
steps to avoid possible rounding errors.
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