# Ex.11.2 Q4 Constructions Solution - NCERT Maths Class 10

## Question

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 \(\begin{align}60^{\circ}\end{align}\).