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?
 A: Your approach is slightly long-winded, since you find the side lengths and then use them to compute the area. Instead, you can compute the area of each triangle $AOB$,$AOC$,$BOC$ (where $O$ is the centre) and add them together. 

In particular, the area of $AOB$ is given by $\frac{1}{2}\cdot 2 \cdot 2 \cdot \sin(\angle AOB)$, and similarly for $AOC$ and $BOC$. Thus the total area is $2 (\sin(\angle AOB) + \sin(\angle AOC) + \sin(\angle BOC))$. 
So, we just need to compute the angles at the centre, which is straightforward.
In particular, using the fact that the angle at the centre is twice the angle at the circumference, the angles at the centre are $90^\circ, 120^\circ, 150^\circ$. These have sines $1, \frac{\sqrt{3}}{2}, \frac{1}{2}$ respectively, so the area is given by $3 + \sqrt{3}$. 
A: Let $A, B, C$ be the vertices of the triangle with opposite sides $a,b,c$ respectively.  Draw $\triangle AOB$ where $O$ is the circumcenter.  Drop the perpendicular to side $c$ from $O$, which meets $c$ at the midpoint $M$ ($\triangle AOB$ is isosceles by constrction).  $\angle AOB$ measures twice $\angle C$ but $OM$ bisects the former angle, thus $\angle AOM$ is congruent with $\angle C$.  So then
$|c|=2|AM|=2R\sin \angle C$, $R$ = circumradius
Likewise
$|a|=2|AM|=2R\sin \angle A$
$|b|=2|AM|=2R\sin \angle B$
This may be rendered as an "extended Law of Sines":
$\frac{\sin \angle A}{|a|}=\frac{\sin \angle B}{|b|}=\frac{\sin \angle C}{|c|}=\frac{1}{2R}$
Now plug-in the area formula
$S = (1/2)|a||b| \sin \angle C =2R^2 \sin \angle A \sin \angle B \sin \angle C$
Then $R=2, |\angle A|=45°, |\angle B|=60°, |\angle C|=180°-45°-60°=75°$.  Put in the respective sine values $1/2, (\sqrt{3})/2, (\sqrt{6}+\sqrt{2})/4$ to get $S=3+\sqrt{3}$.
A: 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: $$\begin{align} A_{\triangle ABC}&=\frac12(2\sqrt2)(\frac{2 \sin 120^\circ}{\sin 30^\circ})\sin75\\ &=\frac{\left(2 \sqrt{2}\right) \left(2 \sqrt{3}\right) \left(\sqrt{3}+1\right)}{\frac{2}{2} 2 \left(2 \sqrt{2}\right)} \\&=\sqrt{3} \left(\sqrt{3}+1\right) \end{align}$$ 
