Given rate of decrease of angle of elevation, calculate the speed of a plane at constant altitude

Here is a practice question for my midterm.. I don't know where in my logic, I'm making a mistake but I'm getting a different answer than I should be getting. Here's how I thought about it So to find the speed of the plane, we find how far it travelled in a minute. i.e. We need to find $x_2 - x_1$. $$x_2 = \frac{5}{tan(\frac{\pi}{6})}$$ $$x_1 = \frac{5}{tan(\frac{\pi}{3})}$$

In the end, I get the answer, $\frac{10}{\sqrt{3}}$

What am I doing wrong? Also, this is for a calculus exam, and I haven't used any calculus here. Am I missing something?

Thanks for the help.

What you are doing wrong is that the instantaneous rate of decrease of the angle is $\frac \pi 6 \frac {\text{rad}}{\text{min}}$. That rate will not continue for the whole minute. You have written the correct equation for $x_1$. Now if $\theta$ is the angle of elevation, you are supposed to say $\frac {d\theta}{dt}=\frac \pi 6$, use the chain rule to relate this to $\frac {dx_1}{dt}$, then compute $\frac {dx_1}{dt}$