A triangle inscribed in a circle of radius $2$ has angles $45^\circ$ and $60^\circ$. What's its area? Points $A$, $B$  and $C$  are on the circumference of a circle with radius 2 such that $\angle BAC = 45$ and $\angle ACB = 60$ . Find the area of $\triangle ABC$ .
I tried using the Law of Sines on the three separate triangles formed  find the side lengths of $AB$, $BC$, and $AC$, and tried to continue with Herons Formula to end up with $(3 + \sqrt{3})/3$, but I got the wrong answer. Any thoughts on a different way to solve this?

The circumcircle of a triangle is centered at the intersection of the perpendicular bisector of its sides
Call this center $O$, and draw $OA, OB, OC$ and effectively you have divided the triangle into three isosceles triangles(because the sides are the radii of the circle). 
Now, we know that the base angles of an isosceles triangle, are equal. So we have this system of equations, (assume angles $A$ is divided into angles $x,y$, $B\to x,z$ and $C\to z,y$): $$x+y=45^\circ\\x+z=60^\circ\\y+z=75^\circ$$ 
Which gives you: $$x\to 15^\circ, y\to 30^\circ, z\to45^\circ$$ 
From here you can workout the interior angles: $$\angle BOA \to 150^\circ, \angle AOC \to 120^\circ, \angle COB\to 90^\circ$$ 
The area of the triangle is defined by: $$A_\triangle=\frac12 a b \sin C$$ 
Looking at $\triangle BOC$, you know that $BC=a=2\sqrt2$ 
Using the law of sines, you get $b$ as: $$\frac{\sin 120^\circ}{b}=\frac{\sin 30^\circ}2\iff b=\frac{2 \sin 120^\circ}{\sin 30^\circ}$$
Now, we can solve for the area as: $$\begin{align} A_{\triangle ABC}&=\frac12(2\sqrt2)(\frac{2 \sin 120^\circ}{\sin 30^\circ})\sin75\\ &=\frac{\left(2 \sqrt{2}\right) \left(2 \sqrt{3}\right) \left(\sqrt{3}+1\right)}{\frac{2}{2} 2 \left(2 \sqrt{2}\right)} \\&=\sqrt{3} \left(\sqrt{3}+1\right) \end{align}$$ 
