22. A triangle with interior angles $60^{\circ}, 45^{\circ}$ and $75^{\circ}$ is inscribed in a circle of radius 2 . What is the area of the triangle?
A $2 \sqrt{3}$
B 4
C $6+\sqrt{3}$
D $6 \sqrt{3}$
E $3+\sqrt{3}$
Solution
E
Let $P, Q$ and $R$ be the vertices of the triangle, let $O$ be the centre of the circle, as shown in the diagram.
The angle subtended by an arc at the centre of a circle is twice the angle subtended at the circumference. Therefore
$$
\begin{array}{l}
\angle Q O R=120^{\circ}, \\
\angle R O P=90^{\circ}
\end{array}
$$
and
$$
\angle P O Q=150^{\circ} \text {. }
$$

We use the notation $[X Y Z]$ for the area of a triangle $X Y Z$.
Using the formula $\frac{1}{2} a b \sin \theta$ for the area of a triangle with sides of lengths $a$ and $b$ with included angle $\theta$, we have
$$
\begin{array}{c}
{[Q O R]=\frac{1}{2}(O Q \times O R) \sin 120^{\circ}=\frac{1}{2}(2 \times 2) \frac{\sqrt{3}}{2}=\sqrt{3},} \\
{[R O P]=\frac{1}{2}(O R \times O P) \sin 90^{\circ}=\frac{1}{2}(2 \times 2)=2}
\end{array}
$$
and
$$
[P O Q]=\frac{1}{2}(O P \times O Q) \sin 150^{\circ}=\frac{1}{2}(2 \times 2) \frac{1}{2}=1 \text {. }
$$

Therefore
$$
[P Q R]=[Q O R]+[R O P]+[P O Q]=\sqrt{3}+2+1=3+\sqrt{3} .
$$