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5.6 What If I Know the Hypotenuse? Pg. 21 Sine and Cosine Ratios

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5.6. What If I Know the Hypotenuse? Pg. 21 Sine and Cosine Ratios. 5.6 – What If I Know the Hypotenuse? Sine and Cosine Ratios. Today you are going to explore two more trigonometric ratios that involve the hypotenuse: Sine and Cosine. Side opposite right angle, longest side. - PowerPoint PPT Presentation

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5.6

What If I Know the Hypotenuse?Pg. 21

Sine and Cosine Ratios

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5.6 – What If I Know the Hypotenuse?Sine and Cosine Ratios

Today you are going to explore two more trigonometric ratios that involve the hypotenuse: Sine and Cosine.

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Hypotenuse:

Opposite Side:

Adjacent Side:

Side opposite right angle, longest side

Hypotenuse

Side opposite angle

Opp

osite

Side touching angle, not the hypotenuse

Adjacent

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O

A

H

sine cosine tangent

sin = opphyp cos = adj

hyp tan = oppadj

S OH C A

H T OA

Note: You cannot use trig ratios on the right angle

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Find the ratio of the indicated angle. Write your answer as a fraction.

O

AH 15

17

8 17

158

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O

A

H

1620 = 4

5

1220 = 3

5

1612 = 4

3

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Like the tangent, your calculator can give you both the sine and cosine ratios for any angle. Locate the "sin" and "cos" buttons on your calculator and use them to find the sine and cosine of 40°. Make sure you get the correct answers and are in degree mode.

sin 40°= _____ cos 40° = _______ tan 40° = _______0.64 0.77 0.84

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Steps to solving with trig:

1.

2.

3.

4.

Label the sides with the reference angle

What side do you know?

What side are you solving for?

Identify which trig ratio to use

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5.31 – TRIGONOMETRYFor each triangle below, decide which side is opposite, adjacent, or the hypotenuse to the given angle. Then determine which of the three trig ratios will help you find x. Write and sole an equation. SOH-CAH-TOA might help.

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O

Hsin 25

9x

1

3.80x

9 sin25x

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H

A cos17 3x1

3.14x

cos17 3x 3

cos17x

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OA tan62 5x

2.66x

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O

H sin34 13

x

x 23.25

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21

21 3

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O

H

sin25 6x

14.2x

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HA cos61 18

x37.13x

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H

A

cos20 10x

9.4x

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5.32 – THE STREETS OF SAN FRANCISCOWhile traveling around the beautiful city of San Francisco, Julia climbed several steep streets. One of the steepest, Filbert Street, has a slope angle of 31.5° according to her guide book. Once Julia finished walking 100 feet up the hill, she decided to figure out how high she had climbed. Julia drew the diagram below to represent the situation.

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a. Can the Pythagorean theorem be used to find the opposite and adjacent side? Why or why not?

No, only know one side

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b. Can special triangles be used to find the opposite and adjacent side? Why or why not?

No, 31.5 isn’t special

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c. Can we use sine, cosine, or tangent to find the opposite and adjacent side? Why or why not?

OH

A

Sine and cosine

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d. Julia still wants to know how many feet she climbed vertically and horizontally when she walked up Filbert Street. Use one of your new trig ratios to find both parts of the missing triangle.

OH

Acos31.5

100x

x 85.26 ft

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OH

Asin31.5

100y

y 52.25 ft

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5.33 – EXACTLY!Martha arrived for her geometry test only to find that she forgot her calculator. She decided to complete as much of each problem as possible.

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a. In the first problem on the test, Martha was asked to find the length of x in the triangle shown at right. Using her algebra skills, she wrote and solved an equation. Her work is shown below. Explain what she did in each step.

Set up equation

multiplied

divided

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b. Martha's answer in part (a) is called an exact answer. Now use your calculator to help Martha find the approximate length of x.

x = 68.62

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c. In the next problem, Martha was asked to find y in the triangle at right. Find the exact answer for y without using a calculator. Then use a calculator to find an approximate value for y.

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OA tan53 5y

1

y 6.64 y 5 • tan53

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Right Triangles Project

Pythagorean Theorem: Given 2 sides45º– 45º– 90º 30º– 60º– 90º

Sine – S

sin-1, cos-1, tan-1

Your NameBlock#

Cosine – C

Tangent – T

2x x x 3 2x x x

OH

AHOA

Clinometer MeasuresArea of Regular Polygons

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Sine – S

O

H

sin θ° = OH

OH

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Cosine – C

A H

cos θ° = AH

AH