Optimization problem: rowing across a lake 
A woman at a point A on the shore of a circular lake with radius $r=3$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time. She can walk at the rate of 10 mph and row a boat at 5 mph. What is the shortest amount of time it would take her to reach point $C$?


My Working
Let $C\hat{A}B=\theta$
$\begin{align*}\therefore AB+arcBC&=\sqrt{3^2+3^2-2\cdot3\cdot3\cdot\cos(\pi - 2\theta)}+3\cdot2\theta\\
&=3\sqrt{2-2\cos(\pi-2\theta)}+6\theta\end{align*}$
$\begin{align*}\therefore \text{time taken} =t=\frac{3\sqrt{2-2\cos(\pi-2\theta)}}{5}+\frac{6\theta}{10}\end{align*}$
To find the point $B$ at which $t$ is minimized, we make $\frac{\operatorname d t}{\operatorname d\theta}=0$:
$\begin{align*}\frac{\operatorname d t}{\operatorname d\theta}=\frac{3-4\sin(\pi-2\theta)}{10\sqrt{2-2\cos(\pi-2\theta)}}+\frac{3}{5}=0\end{align*}$
And the rest of my working gets messy from here. Is there a better method than this?

If you call the center of the circle $O$ and let $\alpha=\angle AOB$, then the woman's rowing distance is $2r\sin \frac{\alpha}{2}=6\sin \frac{\alpha}{2}$; and her walking distance is $r\cdot (\pi-\alpha)=3(\pi-\alpha)$.
So her travelling time is $T(\alpha)=\frac{6}{5}\sin\frac{\alpha}{2}+\frac{3}{10}(\pi-\alpha)$.
I think you will find that this minimizes nicely with calculus methods.
