A jet is flying at its cruising altitude of 6 miles. It’s path carries it directly over Brook, who is observing it and making calculations. At the moment when the elevation angle is 60 degrees, Brook finds that this angle is increasing at 72 degrees per minute. Use this information to calculate the speed of the jet.
Here is what I attempted to do but I did not get the right answer. Any help would be greatly appreciated. Thank you very much
Let's call the angle of view from the ground $\theta$ and the horizontal distance between the airplane and the viewer $x$.
Then, we'd like to find a relation between $x$ and $\theta$. Your idea of drawing a right triangle is a good start. We have the two legs $6$ and $x$, so
$$\tan \theta = \frac{x( \theta )}{6}$$ $$6 \tan \theta = x( \theta )$$
Notice we found a way to express $x$ in terms of $\theta$. Now we want to find the speed which is the derivative of the displacement.
$$x'( \theta ) = 6 \sec^{2} \theta$$
Substitute the required angle of $30 ^ \circ$ (that will be it in your diagram instead of $60 ^ \circ$ because you measure from the horizon. This whole problem can be solved differently with a corresponding right triangle of $60 ^ \circ$)
That will be the instantaneous change of the plane's position when viewed in an angle of $60 ^ \circ$ which is
$$x'( 30 ^ \circ ) = \frac{6}{ \cos ^2 30 ^ \circ } = 8$$
Welcome and well done for your good effort. It's nice that you've showed us exactly what you've tried.
The angle of elevation is measured up from the horizontal. In your diagram, your angle of $60^\circ$ is measured from the vertical. You need to start again with your diagram the other way around and see how you get on.
