Angular rate of change conversion into linear rate of change Studying for a test and there is one single question we couldn't answer, and we have multiple results, no-one knows which one the correct one is.
Image:

D is 1.3m, C is 0.6m and the angle a is 30 degrees. 
Rate of change of a (a') is 150 rad/s. The rate of change of M is asked.
My solution is as follows:
$M=\cos(30^\circ)\cdot0.6+\cos(13.3424)\cdot1.2$
$M'=-(\sin(30^\circ)\cdot0.6\cdot a'+\sin(13.3424)\cdot1.3\cdot B')$
We know that $a'=150$, so we calculate $B'$ ($b=13.3424$)
$\sin(30)\cdot0.6=\sin(b)\cdot1.4$ and thus 
$B'=(\cos(30)\cdot0.6\cdot150)/(\cos(13.3424)\cdot1.3)\,\to\,B'=61.62$ rad/s
Plug into $M'$ formula and $M'=63.485688$ rad/s OR m/s?
Did I do something wrong in my approach? I calculated in radians and converted the angles, does this matter?
 A: Since we wish to find the rate of change of $M$ and we are given rate of change information about $a$ it would be useful to find the relationship between $M$ and $a$ and the fixed parameters $C$ and $D$. This would also allow us to find the answers to the questions regarding specific values of the parameter $a$.
From the Law of Sines and the fact that $b$ will always be an acute angle we see that
\begin{equation}
b=\arcsin\left(\frac{C}{D}\sin a\right)\tag{1}
\end{equation}  
If we let $\mu$ represent the angle opposite $M$ we have
\begin{equation}
M=\frac{D}{\sin a}\cdot\sin\mu\tag{2}
\end{equation} 
But
\begin{eqnarray}
\mu&=&180^\circ-(a+b)\\&=&180^\circ-\left(a+\arcsin\left(\frac{C}{D}\sin a\right)\right)\\
\sin\mu&=&\sin\left(a+\arcsin\left(\frac{C}{D}\sin a\right)\right)\\
&=&\sin a\cos\left(\arcsin\left(\frac{C}{D}\sin a\right)\right)+\cos a\cdot\frac{C}{D}\sin a\\
&=&\frac{\sin a}{D}\sqrt{D^2-C^2\sin^2 a}+\frac{\sin a}{D}\cdot C\cos a
\end{eqnarray}
Substituting this value of $\sin\mu$ into equation $(2)$ gives
\begin{equation}
M=\sqrt{D^2-C^2\sin^2 a}+C\cos a \tag{3}
\end{equation}
Given $D=13,\,C=6, a=30^\circ$ equation $(1)$ gives the result $b=13.3424^\circ$ and equation $(3)$ gives the result $M=4\sqrt{10}+3\sqrt{3}=17.845\,$dm.
To find the rate of change of $M$ we take the time derivative of equation $(3)$ and simplify it to obtain
\begin{equation}
\frac{dM}{dt}=-C\sin a\left(\frac{\cos a}{\sqrt{D^2-C^2\sin^2 a}}+1\right)\cdot\frac{da}{dt}\tag{4}
\end{equation}
Since the radial velocity is already given in terms of "per seconds" it is unnecessary to convert. Substituting all the given values into equation $(4)$ yields an answer of $\frac{dM}{dt}=-480.8$ dm/sec.
