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Question
In the given diagram, O is the centre of the circle. PR and PT are two tangents drawn from the external point P and touching the circle at Q and S respectively. MN is a diameter of the circle. Given ∠PQM = 42° and ∠PSM = 25°.
Find:
- ∠OQM
- ∠QNS
- ∠QOS
- ∠QMS
Solution
a. PR and PT are tangents to the circle with centre O.
Then, ∠OQP = 90°
As, radius is ⊥ to the tangent
Then, ∠OQM = ∠OQP – ∠MQP
= 90° – 42°
= 48°
b. ∠PQM = ∠QNM = 42° ...(By alternate segment theorem)
∠PSM = ∠SNM = 25°
Then ∠QNS = ∠QNM + ∠SNM
= 42° + 25°
= 67°
c. ∠QOS = 2∠QNS ...(since, angle subtended by the arc at the centre is twice the angle subtended by the arc at any other point of the circles.)
= 2 × 67°
= 134°
d. QNSN is a cyclic quadritateral
∠QNS + ∠QMS = 180°
∠QMS = 180° – 67° = 113°
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