Question

An observer stands 25 feet from the base of a flagpole and watches a flag being lowered at a rate of 5 ft/sec. Determine the rate at which the angle of elevation from the observer to the flag is changing at the instant that the flag is 25 feet above eye-level.

Answer 1

Answer:

0.1 rad/s

Step-by-step explanation:

Since the distance from or to the flagpole is not given, I will assume. And my assumption is 50 ft

Now, the elevation angle A = 45 degrees, converting to radians, we have π/4 radians

Remember that the tan of an angle is OPP/HYP, and so

tan A = h/25, on differentiating, we have

d tan A/dA = sec^2 A = (1/25) dh/dA

Next, we have

25 sec^2 A * dA/dt= dh/dt = 5

Making dA/dt the subject of formula, we have

dA/dt = (1/5) cos^2 A

but cos^2 A from trigonometry = 1/2, this means that

dA/dt = .1 radians/second

If you want to convert to degrees, your have

.1 Rad/s( 180 deg/rad) = 18 degrees/second