Copyright © 2009 Pearson Addison-Wesley 7.2-1 7 Applications of Trigonometry and Vectors.

Copyright © 2009 Pearson Addison-Wesley 7.2-1

7Applications of Trigonometry 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

7Applications of Trigonometry and Vectors

Copyright © 2009 Pearson Addison-Wesley

1.1-3

7.2-3

The Ambiguous Case of the Law of Sines

7.2

Description of the Ambiguous Case ▪ Solving SSA Triangles (Case 2) ▪ Analyzing Data for Possible Number of Triangles


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.


1.1-5

7.2-5

If A is acute, there are four possible outcomes.


1.1-6

7.2-6

If A is obtuse, there are two possible outcomes.


1.1-7

7.2-7

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).


1.1-8

7.2-8

Example 1 SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE)

Law of sines(alternative form)

Solve triangle ABC if B = 55°40′, b = 8.94 m, and a = 25.1 m.

Since sin A > 1 is impossible, no such triangle exists.


1.1-9

7.2-9

Example 1 SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE) (continued)

An attempt to sketch the triangle leads to this figure.


1.1-10

7.2-10

Note

In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA).


1.1-11

7.2-11

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:


1.1-12

7.2-12

Example 2 SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued)

Solve separately for triangles


1.1-13

7.2-13



1.1-14

7.2-14



1.1-15

7.2-15

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°.


1.1-16

7.2-16

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.


1.1-17

7.2-17

Example 3 SOLVING THE AMBIGUOUS CASE (ONE TRIANGLE)

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°


1.1-18

7.2-18

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°


1.1-19

7.2-19

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 7.2-1 7 Applications of Trigonometry and Vectors.

Documents

Transcript of Copyright © 2009 Pearson Addison-Wesley 7.2-1 7 Applications of Trigonometry and Vectors.