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 3 Page 665

Recall the definition of tangent.

≈ 14.2 ft

Practice makes perfect

We are given that Annabelle and Rich are setting up decorations for their school dance, and we know that Rich is standing 5 feet directly in front of Annabelle under a disco ball. Let y represent the horizontal distance between Rich and the disco ball.

We know that the angle of elevation from Annabelle to the ball is 40^(∘) and from Rich to the ball is 50^(∘). Since we are asked to evaluate how high the disco ball is, let x represent this height. 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 50^(∘) and tan 40^(∘). tan 50^(∘)=x/y & (I) tan 40^(∘)=x/5+ 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 x in the first equation.
tan50^(∘)=x/y & (I) tan40^(∘)=x/5+y & (II)
ytan50^(∘)=x tan40^(∘)=x/5+y
x=ytan50^(∘) tan40^(∘)=x/5+y
Next we will substitute the value of x into the second equation.
x=ytan50^(∘) tan40^(∘)=x/5+y
x=ytan50^(∘) tan40^(∘)=ytan50^(∘)/5+y
(II):Solve for y
x=ytan50^(∘) (5+y)tan40^(∘)=ytan50^(∘)
x=ytan50^(∘) 5tan40^(∘)+ytan40^(∘)=ytan50^(∘)
x=ytan50^(∘) 5tan40^(∘)=ytan50^(∘)-ytan40^(∘)
x=ytan50^(∘) 5tan40^(∘)=y(tan50^(∘)-tan40^(∘))
x=ytan50^(∘) 5tan40^(∘)/(tan50^(∘)-tan40^(∘))=y
x=ytan50^(∘) y=5tan40^(∘)/(tan50^(∘)-tan40^(∘))
x=ytan50^(∘) y=11.8969...
x=ytan50^(∘) y≈11.9
The value of y is approximately 11.9 feet. By substituting this value into the first equation, we can find the value of x.
x=ytan50^(∘) y≈11.9
x=( 11.9)tan50^(∘) y≈11.9
x=14.1818... y≈11.9
x≈14.2 y≈11.9
The disco ball is approximately 14.2 feet above the ground.