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 5 Page 665

Sketch a diagram describing the given situation. Then recall the definition of the inverse tangent.

≈66^(∘)

Practice makes perfect

We are asked to find the angle of elevation to the peak of a mountain for an observer who is 155 meters from the mountain and whose eyes are 1.5 meters above the ground. We also know that the mountain is 350 meters tall. Let's sketch a diagram describing this situation.

First, let's evaluate the height of the mountain above the observer's line of vision. To do this, we will subtract 1.5 from 350. 350- 1.5=348.5

The height of the mountain above the observer's line of vision is 348.5 meters. Let's add this information to our diagram. Let x represent the angle of elevation.

To evaluate the value of x, we can use one of the inverse trigonometric ratios. Let's start with recalling that the tangent of ∠ A is the ratio of the leg opposite ∠ A to the leg adjacent ∠ A. Using this definition, we can create an equation for tan x. tan x=348.5/155 Next we will recall that if ∠ A is an acute angle and the tangent of A is a, then the inverse tangent of a is the measure of ∠ A. Let's rewrite our equation. tan x=348.5/155 ⇓ x=tan^(-1)348.5/155 Finally, we will solve the equation using a calculator.
x=tan^(-1)348.5/155
x=66.0222...
x≈ 66
The angle of elevation to the peak of the mountain is approximately 66^(∘).