Rate of Change: Angle of Elevation

[emk005]

Homework Statement

A balloon is being tracked from an observation point 0.42 mi from its launch site. Both the launch site and observation point are on level ground. How fast is the angle of elevation to the top of the balloon increasing at the instant it is 20°, if the balloon is rising vertically at a rate of 35 ft/sec. Express the solution in degrees/sec.

Homework Equations

The Attempt at a Solution

.42 mi=2217.6 ft
tanθ=h/b
Differentiated: sec^2(20°) dθ/dt=1/.2217.6 ft dh/dt
sec^2(20°) dθ/dt=1/2217.6 ft (35)
dθ/dt=0.008999

I know the answer is supposed to be .799°/sec but I just can't see where my mistake is. Any help would be appreciated.

 

[SteamKing]

When doing derivatives and integrals of trig functions, the formulas for these quantities only work when the angles are in radians. Convert the 20 degrees to radians to find d(theta)/dt. Once you have the answer, you can convert to degrees / sec.