AP Calculus Related Rates Problem

Question

Can someone help me solve this question? I am a bit confused because there is an angle involved.

Answer 1

Let $\theta$ be the angle of the ladder, and let $\ell$ be the distance between the base of the ladder and the wall. Then, because the ladder forms a right triangle with the wall and floor, $$\cos(\theta) = \frac{\ell}{15}$$ Taking the time derivative of both sides, $$-\frac{d\theta}{dt}\sin(\theta) = \frac{1}{15}\frac{d\ell}{dt}$$ Further, if $h$ is the height of the top of the ladder, then $225=\ell^2+h^2$. Differentiating both sides gives us $$0 = 2\ell\frac{d\ell}{dt}+2h\frac{dh}{dt}$$ Suppose $\ell=9$. Then $h=12$, and since $h'=-2$, we have $\ell'=24/9=8/3$. Further, $\sin(\theta) = 12/15=4/5$. Therefore, using the second equation, we have $\frac{d\theta}{dt} = -\frac{2}{9}$.