Q:

Find the maximum area of a triangle given angle and opposite side

One angle is $30$ degrees and the opposite side is $10$. Two other sides are $x$ and $12$. What is the maximum possible area of given triangle? 
I used law of cosines to find max length of the $x$ then $A = \frac 12\cdot12\cdot x\cdot\sin(30)$ to find the area.  The answer I got was around $55.1769$... is this correct? 
Hopefully you'll understand my question. 

A:

By law of cosines $$100=x^2+144-2\cdot12x\cos30^{\circ}$$ or
$$x^2-12\sqrt3x+44=0,$$ which for the maximum of the area gives
$$x=6\sqrt3+\sqrt{108-44}$$ 0r
$$x=8+6\sqrt3$$ and we obtain:
$$S_{\Delta}=\frac{1}{2}\cdot12\cdot(8+6\sqrt3)\cdot\frac{1}{2}=24+18\sqrt3$$

