Q: AP Calculus Related Rates Problem 
Can someone help me solve this question? I am a bit confused because there is an angle involved.

A: 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 $\frac{d\theta}{dt} = -\frac{2}{9}$.

A: Assume the height of the ladder against the wall is "y" and the distance from the bass of the ladder to the wall is "x."
Assume  = the acute angle between the ground and ladder
We already know dy/dt, so find an equation that will use that simply and keep another variable constant.
sin()=y/15
Now take the derivative of both sides in respect to time (t)
d/dt[sin()] = d/dt[y/15]
 cos()d/dt = (1/15)dy/dt
Now solve for d/dt by dividing over cos() and plugging in variables known.
d/dt = [(1/15)(dy/dt)]/[cos()]
We know the cos() = adj/hyp which at this moment is 9/15
We know dy/dt (given) is -2ft/sec
Plug in and solve to find...
d/dt = -2/9 = -0.222
