Question

a person standing on top of a 10 - foot high sand pile wishes to estimate the width of the pile. he visually locates two rocks on the ground below at the base of the sand pile. the rocks are on opposite sides of the sand pile, and he and the two rocks are in the same vertical plane. if the angles of depression from the top of the sand pile to each of the rocks are 14° and 14°, how far apart are the rocks? round to at least the nearest foot.

Explanation:

Step1: Recall tangent - angle relationship

Let the distance from the base of the sand - pile to each rock be $x$. The height of the sand - pile $h = 10$ feet and the angle of depression $\theta=14^{\circ}$. Since the angle of depression is equal to the angle of elevation from the rock to the top of the sand - pile. We know that $\tan\theta=\frac{h}{x}$.

Step2: Solve for $x$

We have $\tan(14^{\circ})=\frac{10}{x}$, so $x = \frac{10}{\tan(14^{\circ})}$. Using a calculator, $\tan(14^{\circ})\approx0.2493$, then $x=\frac{10}{0.2493}\approx40.11$.

Step3: Find the distance between the two rocks

The two rocks are on opposite sides of the sand - pile. The distance $d$ between the two rocks is $d = 2x$. Substituting the value of $x$, we get $d = 2\times40.11 = 80.22\approx80$ feet.

Answer:

80