Question

a blimp, suspended in the air at a height of 400 feet, lies directly over a line from a sports stadium to a planetarium. if the angle of depression from the blimp to the stadium is 36° and from the blimp to the planetarium is 28°, find the distance between the sports stadium and the planetarium. the distance between the sports stadium and the planetarium is 1302.91 feet. (round to two decimal places as needed.)

Explanation:

Step1: Recall tangent function

The angle of depression is equal to the angle of elevation from the stadium/planetarium to the blimp (alternate interior angles). For a right triangle, $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$. Let the distance from blimp's vertical point to stadium be $x$, to planetarium be $y$. The height of blimp is 400 ft (opposite side for both triangles).

For stadium: $\tan(36^\circ)=\frac{400}{x} \implies x = \frac{400}{\tan(36^\circ)}$

For planetarium: $\tan(28^\circ)=\frac{400}{y} \implies y = \frac{400}{\tan(28^\circ)}$

Step2: Calculate $x$ and $y$

Calculate $\tan(36^\circ)\approx0.7265$, $\tan(28^\circ)\approx0.5317$

$x=\frac{400}{0.7265}\approx550.59$

$y=\frac{400}{0.5317}\approx752.32$

Step3: Total distance

Total distance $d = x + y = 550.59 + 752.32 = 1302.91$

Answer:

1302.91