18. The diagram shows two squares, with sides of length $\frac{1}{2}$, inclined at an angle $2 \alpha$ to one another. What is the value of $x$ ? $\begin{array}{llll}\mathrm{A} \cos \alpha & \mathrm{B} \frac{1}{\cos \alpha} & \mathrm{C} \sin \alpha & \mathrm{D} \frac{1}{\sin \alpha} \quad \mathrm{E} \tan \alpha\end{array}$
Solution: A
Consider the triangle at the bottom of the diagram. We have labelled the points $P, Q$ and $R$, as shown. We let $N$ be the point where the perpendicular from $P$ to $Q R$ meets $Q R$. .
Now, $\angle Q P R=360-90-2 \alpha-90=180-2 \alpha$. In the triangle
$\stackrel{\leftarrow}{Q}$
$P Q R$ we have $P Q=P R=\frac{1}{2}$ as they are sides of the squares. It follows that the triangle $P Q R$ is isosceles and hence $\angle P Q N=\angle P R N=\alpha$. So the triangles $P Q N$ and $P R N$ are congruent. Hence $Q N=N R=\frac{1}{2} x$. Therefore, from the triangle $P Q N$ we have $\cos \alpha=\frac{Q N}{Q P}=\frac{\frac{1}{2} x}{\frac{1}{2}}=x$. That is, $x=\cos \alpha$.