Question

An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 miles per hour. Find the rate at which the angle of elevation tetra is changing when the angle is 30 degrees

Answer 1

Let
angle of elevation=θ
height of plane = H
horizontal distance from observer = x
tan(θ)=x/H

Use implicit differentiation
d(tan(θ))/dt = d(x/H)/dt
sec²(θ)dθ/dt = (dx/dt)/H
dθ/dt=(1/(Hsec²(θ))(dx/dt)
dθ/dt=(cos²(θ)/H)*(dx/dt)
Can you take it from here?

Answer 2

would you just replace H with 5 miles since that's the height the plane is flying at?

Answer 3

This is where you need to do some work.
The formula is general, but the numerical values you substitute have to be consistent in units.

For H=5 miles, and dx/dt=600 miles/hour, it would give dθ/dt in radians/hour.
You need to substitute values in appropriate units of your choice. Finally, you can also change radians to degrees using degrees=radians*180/π.