Draw a circle of radius $4 \, cm$. Construct a pair of tangents to it,the angle between which is $60^{\circ}$. Also,justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

(N/A) Steps of construction:
$1.$ Draw a circle with centre $O$ and radius $4 \, cm$.
$2.$ Draw a radius $OA$ and extend it to $B$ such that $OA = AB = 4 \, cm$. Thus,$OB = 8 \, cm$.
$3.$ Draw a circle with centre $A$ and radius $AO = 4 \, cm$. Let this circle intersect the original circle at points $P$ and $Q$.
$4.$ Join $BP$ and $BQ$. $BP$ and $BQ$ are the required tangents.
Justification:
In $\triangle OAP$,$OA = OP = 4 \, cm$ (radii) and $AP = 4 \, cm$ (radius of circle with centre $A$).
Since all sides are equal,$\triangle OAP$ is an equilateral triangle.
Therefore,$\angle OAP = 60^{\circ}$.
Since $OAB$ is a straight line,$\angle BAP = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
In $\triangle BAP$,$BA = AP = 4 \, cm$. Thus,$\triangle BAP$ is an isosceles triangle.
Therefore,$\angle ABP = \angle APB = (180^{\circ} - 120^{\circ}) / 2 = 30^{\circ}$.
Similarly,$\angle ABQ = 30^{\circ}$.
Thus,$\angle PBQ = \angle ABP + \angle ABQ = 30^{\circ} + 30^{\circ} = 60^{\circ}$.
Measurement:
The distance between the centre $O$ and the point of intersection $B$ is $OB = OA + AB = 4 \, cm + 4 \, cm = 8 \, cm$.