7.3

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Transcript 7.3

Applications of Trigonometry and Vectors

7.3-1

Applications 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

7.3-2

7.3

The Law of Cosines

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

7.3-3

1.1-3

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.

1.1-4 7.3-4

Derivation of the Law of Cosines

Let

ABC

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

The coordinates of

are (

). For angle

, and Thus, the coordinates of

become (

cos

sin

7.3-5

Derivation of the Law of Cosines (continued)

The coordinates of

are (

, 0) and the length of

Using the distance formula, we have Square both sides and expand.

7.3-6

Law of Cosines In any triangle, with sides

, and

1.1-7 7.3-7

Note

= 90 °, then cos

= 0, and the formula becomes the Pythagorean theorem.

1.1-8 7.3-8

Example 1

USING THE LAW OF COSINES IN AN APPLICATION (SAS)

A surveyor wishes to find the distance between two inaccessible points

and

on opposite sides of a lake. While standing at point

, she finds that AC = 259 m,

angle

ACB

= 423 m, and measures 132 °40′. Find the distance

7.3-9

1.1-9

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.

7.3-10

1.1-10

Example 2

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

Solve triangle

ABC b

= 12.9 m, and

= 42.3

= 15.4 m.

°,

B b

since it is opposite the shorter of the two sides and

Therefore,

cannot be obtuse.

7.3-11

1.1-11

Example 2

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

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

≈ 10.47

1.1-12 7.3-12

Caution

If we used the law of sines to find

rather than

, we would not have known whether

is equal to 81.7

° or its supplement, 98.3

°.

1.1-13 7.3-13

Example 3

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

Solve triangle

ABC

a c

= 21.1 ft.

= 9.47 ft,

= 15.9 ft, and Use the law of cosines to find the measure of the largest angle,

. If cos

< 0, angle

is obtuse.

Solve for cos

1.1-14 7.3-14

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

Now find the measure of angle

1.1-15 7.3-15

Example 4

DESIGNING A ROOF TRUSS (SSS)

Find the measure of angle

in the figure.

1.1-16 7.3-16

Four possible cases can occur when solving an oblique triangle.

1.1-17 7.3-17

1.1-18 7.3-18

Heron’s Area Formula (SSS) If a triangle has sides of lengths

, and

1.1-19 7.3-19

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.)

7.3-20

1.1-20

Example 5

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

The semiperimeter

is Using Heron’s formula, the area  is

7.3-21

1.1-21