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 10 Page 666

Recall the definition of tangent.

≈ 7.17 ft

Practice makes perfect

We are given that Claire and Marisa are both waiting to get a rebound during a basketball game and that the height of the basketball hoop is 10 feet. Let y represent the horizontal distance between Claire and the basketball hoop.

We know that the angle of elevation between Claire and the goal is 35^(∘) and the angle of elevation between Marisa and the goal is 25^(∘). Since we are asked to evaluate how far apart the girls are standing, let x represent this distance. Notice that we have two right triangles in our diagram.
To evaluate the lengths of the missing sides, we can use one of the trigonometric ratios. Let's recall that the tangent of ∠ A is the ratio of the leg opposite ∠ A to the leg adjacent ∠ A. Using this definition, we can create the equations for tan 35^(∘) and tan 25^(∘). tan 35^(∘)=10/y & (I) tan 25^(∘)=10/x+ y & (II) As we can see, we need to solve the system of equations. To do this, we can use the Substitution Method. Our first step will be to isolate y in the first equation.
tan35^(∘)=10/y & (I) tan25^(∘)=10/x+y & (II)
ytan35^(∘)=10 tan25^(∘)=10/x+y
y=10/tan35^(∘) tan25^(∘)=10/x+y
y=14.2814... tan25^(∘)=10/x+y
y≈14.28 tan25^(∘)=10/x+y
The value of y is approximately 14.28 feet. By substituting this value into the second equation, we can find the value of x.
y≈14.28 tan25^(∘)=10/x+y
y≈14.28 tan25^(∘)=10/x+ 14.28
Solve for x
y≈14.28 (x+14.28)tan25^(∘)=10
y≈14.28 (x+14.28)=10/tan25^(∘)
y≈14.28 x+14.28=10/tan25^(∘)
y≈14.28 x=10/tan25^(∘)-14.28
y≈14.28 x=7.1650...
y≈14.28 x≈7.17
The girls are standing approximately 7.17 feet far apart.