Triangle in circumference of circle Points $A$, $B$, and $C$ are on the circumference of a circle with radius 2 such that $\angle BAC = 45^\circ$ and $\angle ACB = 60^\circ$. Find the area of $\triangle ABC$.
How would I start this problem? I think it has to do with the extended law of sines, but I'm not sure.
Thanks

Use the sines law: $\frac{AB}{sin(C)}=\frac{AC}{sin(B)}=\frac{BC}{sin(A)}=2R=2\cdot 2=4$.  So:
$$AB=4 sin(60^ {\circ})=2 \sqrt{3} $$
$$ BC=4sin(45^{\circ})=2 \sqrt{2} $$
Then: $S_{ \triangle ABC}=\frac{AB \cdot BC \cdot \sin (\widehat{ABC})}{2}=\frac{2\sqrt{2}\cdot 2\sqrt{3} \cdot \sin(75^{ \circ})}{2}=2\sqrt{6}\cdot \frac{1+\sqrt{3}}{2 \sqrt{2}}=\sqrt{3}+3$.
