Draw a circle of radius $4\ cm$. Draw two tangents to the circle inclined at an angle of $60^{o}$ to each other.

Given: Radius of the circle$=4\ cm$.

To do: To draw a circle of the given radius and to draw two tangents to the circle inclined at an angle of $60^{o}$ to each other.

$( i)$. Take a point O on the plane of the paper and draw a circle of radius $ OA=4\ cm$.

$( ii)$. Produce OA to B such that $OA=AB=4\ cm$.

$( iii)$. Draw a circle with center at A and radius AB.

$( iv)$. Suppose it cuts the circle drawn in step $( i)$ at P and Q.

$( v)$. Join $BP$ and $BQ$ to get the desired tangents.

In $\vartriangle OAP,\ OA\ =\ OP\ =\ 4\ cm\ ...\ ( radii\ of\ the\ same\ circle)$

Also, $AP\ =\ 4\ cm\ \dotsc .( Radius\ of\ the\ circle\ with\ center\ A)$

$\therefore \ \vartriangle OAP\ is\ equilateral.$

$\therefore \ \angle PAO\ =\ 60^{o}$

$\therefore \ \angle BAP\ =\ 120^{o}$

In $\vartriangle BAP$, we have $BA\ =\ AP$ and $\angle \ BAP\ =\ 120^{o}$

$\therefore \ \angle ABP\ =\ \angle APB\ =\ 30^{o}$

Similarly we can get $\angle ABQ\ =\ 30^{o}$

$\therefore \ \angle PBQ\ =\ 60^{o}$