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In Fig. 10.37, /P Q R\ =\ 100^@, where ...

In Fig. 10.37, `/_P Q R\ =\ 100^@`, where P, Q and R are points on a circle with centre O. Find `/_O P Rdot`

Text Solution

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Since all points `P, Q, R, S` lie on the circle, `PQRS` becomes a cyclic quadrilateral.
We know that the sum of either pair of opposite angles of a cyclic quadrilateral is `180^@`
Therefore,`anglePQR + anglePSR = 180^@`
`100^@ + anglePSR = 180^@`
`anglePSR = 180^@ - 100^@ = 80^@`
We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle
Therefore,`anglePOR = 2anglePSR`
= `2 × 80^@`
= `160^@`
Consider the `triangleOPR`. It is an isosceles triangle as `OP = OR` = Radius of the circle.
Thus, `angleOPR = angleORP`
Sum of all angles in a triangle is `180^@`
Therefore, `angleOPR + anglePOR + angleORP = 180^@`
`angleOPR + 160^@ + angleOPR = 180^@`
`2angleOPR = 180^@ - 160^@`
`angleOPR = 10^@`
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