Question

an airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highway on opposite sides of the plane. the angle of depression to one car is 34°, and that to the other is 57°. how far apart are the cars? (round your answer to the nearest foot.)

Explanation:

Step1: Define variables for distances

Let $x_1$ be the distance from the point on the highway directly below the plane to the first car and $x_2$ be the distance to the second car. The elevation of the plane $h = 5150$ ft.
We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. For the angle of depression of $34^{\circ}$, the angle of elevation from the car to the plane is also $34^{\circ}$. So $\tan34^{\circ}=\frac{h}{x_1}$, then $x_1=\frac{h}{\tan34^{\circ}}$. For the angle of depression of $57^{\circ}$, the angle of elevation from the car to the plane is $57^{\circ}$, and $\tan57^{\circ}=\frac{h}{x_2}$, then $x_2 = \frac{h}{\tan57^{\circ}}$.

Step2: Calculate $x_1$

$x_1=\frac{5150}{\tan34^{\circ}}$. Since $\tan34^{\circ}\approx0.6745$, $x_1=\frac{5150}{0.6745}\approx7635.3$.

Step3: Calculate $x_2$

$x_2=\frac{5150}{\tan57^{\circ}}$. Since $\tan57^{\circ}\approx1.5399$, $x_2=\frac{5150}{1.5399}\approx3343.0$.

Step4: Find the distance between the cars

The distance $d$ between the two cars is $d=x_1 + x_2$.
$d\approx7635.3+3343.0 = 10978.3\approx10978$ ft.

Answer:

$10978$