Question

1. a vertical radio tower is 645 feet high above level ground. how far would you need to be from the base of the tower so that the angle of elevation to the top of the tower would be 35.0°? (the “angle of elevation” is the angle measured from level ground up to the top of the tower.)

Explanation:

Step1: Set up the tangent - ratio equation

We know that in a right - triangle (where the radio tower is the vertical side, the distance from the base of the tower is the horizontal side, and the line of sight to the top of the tower is the hypotenuse), the tangent of the angle of elevation is given by $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Let $h = 645$ feet be the height of the tower (opposite side) and $d$ be the distance from the base of the tower (adjacent side), and $\theta = 35.0^{\circ}$. So, $\tan\theta=\tan(35.0^{\circ})=\frac{h}{d}$.

Step2: Solve for $d$

We can re - arrange the equation $\tan(35.0^{\circ})=\frac{h}{d}$ to get $d=\frac{h}{\tan(35.0^{\circ})}$. Substitute $h = 645$ feet and $\tan(35.0^{\circ})\approx0.7002$. Then $d=\frac{645}{0.7002}\approx921$ feet.

Answer:

Approximately 921 feet