Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ \frac{4}{\sqrt{3}} }$
Triangle $OAB$ has the property that $\angle O=30^{\circ}$ and $AB=1$. 
From the [Law of Sines](https://artofproblemsolving.com/wiki/index.php/Law_of_Sines), $\frac{\sin{\angle A}}{OB}=\frac{\sin{\angle O}}{AB}$. 
Since $\sin 30^\circ = \frac{1}{2}$, we have:
$\frac{\sin{\angle A}}{OB}=\frac{\frac{1}{2}}{1}$
$2\sin{\angle A}=OB$. 
The maximum of $\sin{\angle A}$ is $1$ when $\angle A = 90^\circ$, so the maximum of $OB$ is $2\Rightarrow \mathrm{(D)}$.