Differentiation problem - airplane/observer question

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.

• Regarding the use of formulas, check out meta.math.stackexchange.com/questions/5020/…. I've edited your answer to add this. You can view it by either right clicking something and choosing "Show Math As">"TeX Commands" or choosing the "edit" option for the post and looking at the markup. – RandomUser May 23 '14 at 20:14
• @RandomUser Thank you! – Badalyan May 23 '14 at 22:35

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.