Top Banner
Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice
16

Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Dec 25, 2015

Download

Documents

Dustin Nichols
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Geometry

One is always a long way from solving a problem until one actually has the answer.

Stephen Hawking

Today:• 9.5

Instruction

• Practice

Page 2: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Geometry - Yesterday

One is always a long way from solving a problem until one actually has the answer.

Stephen Hawking

Assignment:• Trig Notes

Page 3: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

9.5 Trigonometric Ratios

Objectives:

1. Find the sine, cosine and tangent of acute angles.

2. Use trig ratios in real life. 

Vocabulary: Sine, Cosine, Tangent, Angle of Elevation

Page 4: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Finding Trig Ratios

A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The word trigonometry is derived from the ancient Greek language and means measurement of triangles. The three basic trigonometric ratios are sine, cosine, and tangent, which are abbreviated as sin, cos, and tan respectively.

Page 5: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Trigonometric Ratios

Let ∆ABC be a right triangle. The since, the cosine, and the tangent of the acute angle A are defined as follows.

ac

bside adjacent to angle A

Sideoppositeangle A

hypotenuse

A

B

C

sin A =Side opposite A

hypotenuse= a

c

cos A =Side adjacent A

hypotenuse= b

c

tan A =Side opposite A

Side adjacent A

= a

b

Page 6: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Ex. 1: Finding Trig Ratios

Compare the sine, the cosine, and the tangent ratios for A in each triangle beside.

By the SSS Similarity Theorem, the triangles are similar. Their corresponding sides are in proportion which implies that the trigonometric ratios for A in each triangle are the same.

15

817

A

B

C

7.5

48.5

A

B

C

Page 7: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Ex. 1: Finding Trig Ratios

15

817

A

B

C

7.5

48.5

A

B

C

Large Small

sin A = opposite

hypotenuse

cosA = adjacent

hypotenuse

tanA = opposite

adjacent

8

17≈ 0.4706

15

17≈ 0.8824

8

15≈ 0.5333

4

8.5≈ 0.4706

7.5

8.5≈ 0.8824

4

7.5≈ 0.5333

Trig ratios can be expressed as fractions or decimals.

Page 8: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Finding Trig Ratios - On Your OWN

15

817

A

B

C

sin B = opposite

hypotenuse

cosB = adjacent

hypotenuse

tanB = opposite

adjacent

Take the big triangle from example #1, now focus on angle B.

How does this change your labels?

Page 9: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Using a Calculator

You can use a calculator to approximate the sine, cosine, and the tangent of 74. Make sure that your calculator is in degree mode. The table shows some sample keystroke sequences accepted by most calculators.

Page 10: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Sample keystrokes

Sample keystroke sequences

Sample calculator display Rounded

Approximation

74

74

0.961262695 0.9613

0.275637355 0.2756

3.487414444 3.4874

sinsin

ENTER

74

74

COS

COS

ENTER

74

74

TAN

TANENTER

Page 11: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Using Trigonometric Ratios in Real-life

Suppose you stand and look up at a point in the distance. Maybe you are looking up at the top of a tree as in Example 6. The angle that your line of sight makes with a line drawn horizontally is called angle of elevation.

Page 12: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Indirect Measurement You are measuring the

height of a Sitka spruce tree in Alaska. You stand 45 feet from the base of the tree. You measure the angle of elevation from a point on the ground to the top of the top of the tree to be 59°. To estimate the height of the tree, you can write a trigonometric ratio that involves the height h and the known length of 45 feet.

Page 13: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

The math

tan 59° =opposite

adjacent

tan 59° =h

4545 tan 59° = h

45 (1.6643) ≈ h

74.9 ≈ h

Write the ratio

Substitute values

Multiply each side by 45

Use a calculator or table to find tan 59°

Simplify

The tree is about 75 feet tall.

Page 14: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Estimating Distance

Escalators. The escalator at the Wilshire/Vermont Metro Rail Station in Los Angeles rises 76 feet at a 30° angle. To find the distance d a person travels on the escalator stairs, you can write a trigonometric ratio that involves the hypotenuse and the known leg of 76 feet.

d76 ft

30°

Page 15: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Now the math

d76 ft

30°

sin 30° =oppositehypotenuse

sin 30° =76

d

d sin 30° = 76

sin 30° 76

d =

0.5 76

d =

d = 152

Write the ratio for sine of 30°

Substitute values.

Multiply each side by d.

Divide each side by sin 30°

Substitute 0.5 for sin 30°

Simplify

A person travels 152 feet on the escalator stairs.

Page 16: Geometry One is always a long way from solving a problem until one actually has the answer. Stephen Hawking Today: 9.5 Instruction Practice.

Geometry

One is always a long way from solving a problem until one actually has the answer.

Stephen Hawking

Assignment:

9.5 p 562 #9-41 Every Other Odd