20. The diagram shows a semicircle of radius 1 inside an isosceles triangle. The diameter of the semicircle lies along the 'base' of the triangle, and the angle of the triangle opposite the 'base' is equal to $2 \theta$. Each of the two equal sides of the triangle is tangent to the semicircle.
What is the area of the triangle?
A $\frac{1}{2} \tan 2 \theta$
B $\sin \theta \cos \theta$
C $\sin \theta+\cos \theta$
D $\frac{1}{2} \cos 2 \theta$
$\mathrm{E} \frac{1}{\sin \theta \cos \theta}$
Solution
E
Let the vertices of the triangle be $P, Q$ and $R$, with $P Q=P R$, as shown in the diagram, and let $O$ be the midpoint of $Q R$. We also let $T$ be the point where the semicircle touches $P R$. In the triangles $P Q O$ and $P R O$, we have $P Q=P R, Q O=R O$ and the side $P O$ is common. Therefore the triangles are congruent (SSS). Hence the area of the triangle $P Q R$ is twice the area of the triangle $P O R$.

It also follows that $\angle Q P O=\angle R P O=\theta$. Hence $P O$ is the bisector of $\angle Q P R$. Hence $O$ is equidistant from $P Q$ and $P R$ and is therefore the centre of the semicircle.

We can also deduce that $\angle P O Q=\angle P O R=90^{\circ}$.
$O T$ is a radius of the semicircle and therefore has length 1 and is perpendicular to the tangent $P R$. Hence $P T O$ is a right-angled triangle and therefore
$$
\frac{O T}{O P}=\sin \theta
$$
and hence
$$
O P=\frac{O T}{\sin \theta}=\frac{1}{\sin \theta} .
$$

Also, from the triangle $P O R$, we have
$$
\frac{O P}{P R}=\cos \theta
$$
and hence
$$
P R=\frac{O P}{\cos \theta}=\frac{1}{\sin \theta \cos \theta} .
$$

If we regard $P R$ as the base of the triangle $P O R$, and $O T$ as its base, we see that
$$
\text { area of } P O R=\frac{1}{2}(P R \times O T)=\frac{1}{2 \sin \theta \cos \theta} \text {. }
$$

It follows that the area of the triangle $P Q R$ is $\frac{1}{\sin \theta \cos \theta}$.