An airplane is flying at a constant speed at a constant altitude of $3$ km in a straight line that will take it directly over an observer at a ground level.
At a given instant the observer noted that that the angle $\theta$ from the ground to the plane is $\frac{\pi}{3}$ radians and is increasing at $\frac{1}{60}$ radians per second. Find the speed, in km/h, at which the airplane is moving towards the observer.
In my calculations, I found that the horizontal distance between the guy and the airplane is $1.732$ km. So $$\frac{d\theta}{dt} = \frac{1}{60} = \frac{d\theta}{ds} \cdot \frac{ds}{dt}$$
Is this the right way of doing it?
P.S. Also, it would be great if someone would teach me how to use formulas instead of text in this website, as I am new here.
I have the following. Yes the initial horizontal distance is $s=3*\cot(\pi/3)=\sqrt{3}$. Now $s-ds=3*\cot(\pi/3+d\theta)$. So $$ {ds\over dt}=-3\cot(\theta)^\prime {d\theta \over dt}=-3(\cot(\theta)^2-1) {d\theta \over dt} $$ So if you are looking for the initial speed then it will be $$ {ds\over dt}=-3(1/3-1)/60=1/30 km/sec=120 km/h $$ I hope I did not make any mistake here.