Copyright © 2009 Pearson Addison-Wesley 7.3-1 7 Applications of Trigonometry and Vectors.

Copyright © 2009 Pearson Addison-Wesley 7.3-1

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

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7.3-3

The Law of Cosines7.3Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle


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Triangle Side Length Restriction

In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.


Derivation of the Law of Cosines

Let ABC be any oblique triangle located on a coordinate system as shown.

The coordinates of A are (x, y). For angle B,

and

Thus, the coordinates of A become (c cos B, c sin B).


Derivation of the Law of Cosines (continued)

The coordinates of C are (a, 0) and the length of AC is b.

Using the distance formula, we have

Square both sides and expand.


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Law of Cosines

In any triangle, with sides a, b, and c,


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7.3-8

Note

If C = 90°, then cos C = 0, and the formula becomes the Pythagorean theorem.


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7.3-9

Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS)

A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that AC = 259 m, BC = 423 m, and angle ACB measures 132°40′. Find the distance AB.


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Example 1 USING THE LAW OF COSINES IN AN APPLICATION (SAS)

Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle.

The distance between the two points is about 628 m.


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Example 2 USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SAS)

Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m.

B < C since it is opposite the shorter of the two sides b and c. Therefore, B cannot be obtuse.


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Example 2 USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SAS) (continued)

≈ 10.47

Use the law of sines to find the measure of another angle.

Now find the measure of the third angle.


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Caution

If we used the law of sines to find C rather than B, we would not have known whether C is equal to 81.7° or its supplement, 98.3°.


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Example 3 USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SSS)

Solve triangle ABC if a = 9.47 ft, b = 15.9 ft, and c = 21.1 ft.

Use the law of cosines to find the measure of the largest angle, C. If cos C < 0, angle C is obtuse.

Solve for cos C.


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Example 3 USING THE LAW OF COSINES TO SOLVE A TRIANGLE (SSS) (continued)

Use either the law of sines or the law of cosines to find the measure of angle B.

Now find the measure of angle A.


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Example 4 DESIGNING A ROOF TRUSS (SSS)

Find the measure of angle B in the figure.


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Four possible cases can occur when solving an oblique triangle.


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Heron’s Area Formula (SSS)

If a triangle has sides of lengths a, b, and c, with semiperimeter

then the area of the triangle is


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Example 5 USING HERON’S FORMULA TO FIND AN AREA (SSS)

The distance “as the crow flies” from Los Angeles to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of Earth.)


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Example 5 USING HERON’S FORMULA TO FIND AN AREA (SSS) (continued)

The semiperimeter s is

Using Heron’s formula, the area is

Copyright © 2009 Pearson Addison-Wesley 7.3-1 7 Applications of Trigonometry and Vectors.

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