A hot air ballon lifts off the earth 300m away from an observer and rises straight up at a rate of 80m/min. At what rate is the angle of inclination of the observers line of sight increasing at the time the balloon is 300, above the ground?

Answer 1

when the balloon is at height h,

tanθ = h/300
so,

sec^2θ dθ/dt = 1/300 dh/dt

you know that when h=300, θ=π/4.
You know dh/dt = 80, so just solve for dθ/dt