McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Angles of Elevation and Depression
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Exercise 20 Page 667

Practice makes perfect
a We are given that Enrique and Trina are at the topmost point on a Ferris wheel that has a diameter of 100 meters, and we're asked to evaluate the distance between them and their friend India. Let's sketch a diagram describing this situation. We also know that the angle of depression is 57^(∘).
Recall that the angle of depression and the angle of elevation are congruent. Let x represent the distance we are looking for.
To evaluate the value of x, we can use one of the trigonometric ratios. Let's start with recalling that the sine of ∠ A is the ratio of the leg opposite ∠ A to the hypotenuse. Using this definition, we can create an equation for sin57^(∘). sin 57^(∘)=100/x Now we will solve the equation using a calculator.
sin 57^(∘)=100/x
xsin57^(∘)=100
x=100/sin57^(∘)
x=119.236...
x≈ 119.2
They are approximately 119.2 meters from India.
b In this part we are given that the 80-meter cliff-dropping ride has an angle of depression of 46^(∘), and we're asked to determine the length of the slide, which we will call l. Let's sketch a diagram describing this situation.
To evaluate the value of l, we can use one of the trigonometric ratios. Let's again recall that the sine of ∠ A is the ratio of the leg opposite ∠ A to the hypotenuse. Using this definition, we can create an equation for sin46^(∘). sin 46^(∘)=80/l Now we will solve the equation using a calculator.
sin 46^(∘)=80/l
lsin46^(∘)=80
l=80/sin46^(∘)
l=111.213...
l≈ 111.2
The length of the slide is approximately 111.2 meters.