Transcript 7.2

7
Applications
of
Trigonometry
and Vectors
Copyright © 2009 Pearson Addison-Wesley
7.2-1
Applications of Trigonometry
7 and Vectors
7.1 Oblique Triangles and the Law of Sines
7.2 The Ambiguous Case of the Law of
Sines
7.3 The Law of Cosines
7.4 Vectors, Operations, and the Dot
Product
7.5 Applications of Vectors
Copyright © 2009 Pearson Addison-Wesley
7.2-2
7.2 The Ambiguous Case of the
Law of Sines
Description of the Ambiguous Case ▪ Solving SSA Triangles
(Case 2) ▪ Analyzing Data for Possible Number of Triangles
Copyright © 2009 Pearson Addison-Wesley
1.1-3
7.2-3
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.
Copyright © 2009 Pearson Addison-Wesley
7.2-4
If A is acute, there are four possible outcomes.
Copyright © 2009 Pearson Addison-Wesley
1.1-5
7.2-5
If A is obtuse, there are two possible outcomes.
Copyright © 2009 Pearson Addison-Wesley
1.1-6
7.2-6
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).
Copyright © 2009 Pearson Addison-Wesley
1.1-7
7.2-7
Example 1
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.
Copyright © 2009 Pearson Addison-Wesley
1.1-8
7.2-8
Example 1
SOLVING THE AMBIGUOUS CASE (NO
SUCH TRIANGLE) (continued)
An attempt to sketch the triangle leads to this figure.
Copyright © 2009 Pearson Addison-Wesley
1.1-9
7.2-9
Note
In the ambiguous case, we are given
two sides and an angle opposite one of
the sides (SSA).
Copyright © 2009 Pearson Addison-Wesley
1.1-10
7.2-10
Example 2
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:
Copyright © 2009 Pearson Addison-Wesley
1.1-11
7.2-11
Example 2
SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
Solve separately for triangles
Copyright © 2009 Pearson Addison-Wesley
1.1-12
7.2-12
Example 2
Copyright © 2009 Pearson Addison-Wesley
SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
1.1-13
7.2-13
Example 2
Copyright © 2009 Pearson Addison-Wesley
SOLVING THE AMBIGUOUS CASE (TWO
TRIANGLES) (continued)
1.1-14
7.2-14
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°.
Copyright © 2009 Pearson Addison-Wesley
1.1-15
7.2-15
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.
Copyright © 2009 Pearson Addison-Wesley
1.1-16
7.2-16
SOLVING THE AMBIGUOUS CASE (ONE
TRIANGLE)
Example 3
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°
Copyright © 2009 Pearson Addison-Wesley
1.1-17
7.2-17
Example 3
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°
Copyright © 2009 Pearson Addison-Wesley
1.1-18
7.2-18
Example 4
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.
Copyright © 2009 Pearson Addison-Wesley
1.1-19
7.2-19