Draw a pair of tangents to a circle of radius $5\, cm$ which are inclined to each other at an angle of $60^{\circ}$. Also,provide the justification for the construction.

(N/A) The tangents can be constructed in the following manner:
$1.$ Draw a circle of radius $5\, cm$ with center $O$.
$2.$ Take a point $A$ on the circumference of the circle and join $OA$. Draw a perpendicular to $OA$ at point $A$.
$3.$ Draw a radius $OB$ making an angle of $120^{\circ} (180^{\circ} - 60^{\circ})$ with $OA$.
$4.$ Draw a perpendicular to $OB$ at point $B$. Let both the perpendiculars intersect at point $P$. $PA$ and $PB$ are the required tangents inclined at an angle of $60^{\circ}$.
Justification:
The construction can be justified by proving that $\angle APB = 60^{\circ}$.
By our construction:
$\angle OAP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
$\angle OBP = 90^{\circ}$ (Tangent is perpendicular to the radius at the point of contact)
And $\angle AOB = 120^{\circ}$
We know that the sum of all interior angles of a quadrilateral is $360^{\circ}$.
In quadrilateral $OAPB$:
$\angle OAP + \angle AOB + \angle OBP + \angle APB = 360^{\circ}$
$90^{\circ} + 120^{\circ} + 90^{\circ} + \angle APB = 360^{\circ}$
$300^{\circ} + \angle APB = 360^{\circ}$
$\angle APB = 360^{\circ} - 300^{\circ} = 60^{\circ}$
This justifies the construction.