21. The diagram shows ten equal discs that lie between two concentric circles - an inner circle and an outer circle. Each disc touches two neighbouring discs and both circles. The inner circle has radius 1 .
What is the radius of the outer circle?
A $2 \tan 36^{\circ}$
B $\frac{\sin 36^{\circ}}{1-\sin 36^{\circ}}$
C $\frac{1+\sin 18^{\circ}}{1-\sin 18^{\circ}}$
D $\frac{2}{\cos 18^{\circ}}$
E $\frac{9}{5}$
Solution
C
Let $O$ be the common centre of the inner and outer circles, let $C$ be the centre of one of the discs and let $T$ be the point where one of the tangents from $O$ meets the disc, as shown in the diagram.

Let the radius of each of the discs be $r \mathrm{~cm}$. We see from the diagram that the radius of the outer circle is $(1+2 r) \mathrm{cm}$.

There are ten equal discs surrounding the inner circle. Therefore the angle between the two tangents from $O$ to one of the discs is $\frac{1}{10}\left(360^{\circ}\right)$, that is, $36^{\circ}$.
Since $\angle T O C$ is one half of this angle, we have $\angle T O C=18^{\circ}$.
The tangent $O T$ is at right angles to the radius $C T$. Therefore $T O C$ is a right-angled triangle in which the hypotenuse, $O C$, has length $(1+r) \mathrm{cm}$, and the side opposite the angle of $18^{\circ}$ has length $r \mathrm{~cm}$.
It follows that
$$
\sin 18^{\circ}=\frac{r}{1+r}
$$

From this equation we deduce that
$$
(1+r) \sin 18^{\circ}=r
$$
and hence that
$$
\sin 18^{\circ}=\left(1-\sin 18^{\circ}\right) r
$$

It follows that
$$
r=\frac{\sin 18^{\circ}}{1-\sin 18^{\circ}} .
$$

Therefore the radius of the outer circle is given by
$$
1+2 r=1+\frac{2 \sin 18^{\circ}}{1-\sin 18^{\circ}}=\frac{\left(1-\sin 18^{\circ}\right)+2 \sin 18^{\circ}}{1-\sin 18^{\circ}}=\frac{1+\sin 18^{\circ}}{1-\sin 18^{\circ}}
$$