Ex.11.2 Q4 Constructions Solution - NCERT Maths Class 10

Draw a pair of tangents to a circle of radius \(5 \,\rm{cm}\) which are inclined to each other at an angle of \(60^\circ\).

Text Solution

Steps:

Steps of construction:

(i)    With \(O\) as centre and \(5 \,\rm{cm}\) as radius draw a circle.

(ii)    Take a point \(A\) on the circumference of the circle and join \(OA\).

(iii)   Draw \(AX\) perpendicular to \(OA\).

(iv)   Construct \(\begin{align}\angle \rm{AOB}=120^{\circ} \end{align}\) where \(B\) lies on the circumference.

(v)    Draw \(BY\) perpendicular to \(OB\).

(vi)   Both \(AX\) and \(BY\) intersect at \(P\).

(vii)  \(PA\) and \(PB\) are the required tangents inclined at \(\begin{align}60^{\circ}\end{align}\) .

Proof:

\(\angle {{OAP}} = \angle {{OBP}} = 90^\circ\) (By construction)

\(\angle {{AOB}} = 120^\circ\) (By construction)

In quadrilateral \(OAPB\),

\[\begin{align} \angle {{APB}}& = 360^\circ - [\angle {{OAP}} + \angle {{OBP}} + \angle {{AOB}}]\\ &= 360^\circ - [90^\circ + 90^\circ + 120^\circ ]\\ &= 360^\circ - 300^\circ \\ &= 60^\circ \end{align}\]

Hence \(PA\) and \(PB\) are the required tangents inclined at \(\begin{align}60^{\circ}\end{align}\).