5.8

What If I Know the Hypotenuse?

Pg. 23Sine and Cosine Ratios

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

In the previous section, you used the idea of similarity in right triangles to find a relationship between the acute angles and the lengths of the legs of a right triangle. However, we do not always work just with the legs of a right triangle – sometimes we only know the length of the hypotenuse. By the end of this lesson, you will be able to use two new trigonometric ratios that involve the hypotenuse of right triangles.

5.43 – MISSING THE LEGS

a. What is the relationship between the two triangles below? Explain your reasoning.

Similar, AA~

b. Find the rise for the slope triangles shown at right.

x2 + 2.32 = 32

x2 + 5.29 = 9

x2 = 3.71x = 1.93

1.93

3.86

adjacent

hypotenusec. Find the ratio for each triangle.

Why must these ratios be equal?

1.93

3.86

OA

H

O

A

H2.33

4.66

= 2330

= 2330

Ratios are equal because triangles are

similar

5.44 – MISSING THE LEGS In the last problem, you used a ratio that included the hypotenuse of ∆ABC. One of these ratios is known as the sine ratio (pronounced "sign"). This is the ratio of the length of the side opposite the acute angle to the length of the hypotenuse.

sin

opposite

hypotenuse

O

A

H

cos

adjacent

hypotenuse

O

A

H

5.45 – CALCLUATOR TIME!a. 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° = _____________________

0.64

0.77

Shows each ratio in relation to the sides

b. Use a trig ratio to write an equation and solve for a in the diagram at right. Does this require the sine ratio or the cosine ratio? Begin by labeling the opposite, adjacent, and hypotenuse. What side do you know? What side are you solving for?

OH

sin23 15

a

1

5.86a 15 sin23a A

c. Likewise, write an equation and solve for b in the triangle at right.

OHcos37

8

b1

6.39b

8 cos37b

A

5.46 – 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.

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

No, only know one side

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

No, 31.5 isn’t special

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

OH

A

Sine and cosine because we know the hypotenuse and one angle

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

A

cos31.5 100

x1

x 85.26 ft100 cos31.5x

OH

Asin 31.5

100

y

1

y 52.25 ft100 sin31.5y

5.47 – 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.

O

HA sin 25

9

x

1

3.80x

9 sin 25x

OH

A cos17 3

x1

3.14x

cos17 3x 3

cos17x

5

O

H

A tan62 5

x1

2.66x

tan62 5x 5

tan62x

O

H

A

sin34 13

x1

x 23.25

x sin34 13

x

13

sin34

21

21 3

OH

Asin28

19

x

1

8.92x 19 sin28x

O

H

A sin25 6

x1

14.2x

6

sin 25x

sin 25 6x

O

HA cos61 18

x1

37.13x

18

cos61x

cos61 18x

16

8 3

OH

A

cos20 10

x1

9.4x

10 cos20x

8

5.48 – 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.

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

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

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.

O

H

A tan53 5

y1

6.64x 5 tan53x

5.49 – CONCLUSIONSDescribe how to determine when to use the sine, cosine, and tangent ratios. What are the steps in setting up an equation? How can you write an exact answer?