# Ex.10.5 Q3 Circles Solution - NCERT Maths Class 9

## Question

In the given figure, \(\begin {align} \angle {PQR}=100^{\circ} \end {align}\) where \(P, Q\) and \(R\) are points on a circle with center \(O\) . Find \(\begin {align}\angle {OPR} \end {align}\).

## Text Solution

**What is known?**

**\(3\)** points on the circle and

\(\begin {align}\angle {PQR}=100^{\circ} \end {align}\)

**What is unknown?**

Value of \(\angle {PQR} \)

**Reasoning:**

- The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
- A quadrilateral \(\begin {align} {ABCD} \end {align}\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^{\circ}. \end {align}\)

**Steps:**

Mark any point on the major arc side (opposite side to point \(\begin {align} {Q}\end {align}\)) as \(\begin {align} {S.}\end {align}\)

Since all points \(\begin {align} {P, Q, R, S}\end {align}\) lie on the circle, \(\begin {align} {PQRS}\end {align}\) becomes a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^\circ.\end {align}\)

Therefore,

\[\begin{align} \angle {PQR}+\angle {PSR} &=180^{\circ} \\ 100^{\circ}+\angle {PSR} &=180^{\circ} \\ \angle {PSR} &=180^{\circ}-100^{\circ} \\ &=80^{\circ} \end{align}\]

We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Therefore,

\[\begin{aligned} \angle {POR} &=2 \angle {PSR} \\ &=2 \times 80^{\circ} \\ &=160^{\circ} \end{aligned}\]

Consider the \(\begin{aligned} \Delta {OPR.}\end{aligned}\) It is an isosceles triangle as

\[\begin{align} OP &= OR \\ &= \text{Radius of the circle.}\end{align}\]

\[\begin {align}∴ \angle {OPR}=\angle {ORP} \end {align}\]

Sum of all angles in a triangle is \(\begin {align}180^{\circ}. \end {align}\)

Therefore,

\[\begin{align}\angle {OPR}+ \!\angle {POR}+ \! \angle {ORP}&=180^{\circ} \\ \angle {OPR}+160^{\circ}+\angle {OPR}&=180^{\circ} \end{align}\]

\[\begin{align} 2 \angle {OPR}&= 180^{\circ} -160^{\circ} \\ \angle {OPR}&=10^{0}\end{align}\]