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:
(a) ∠OQM
(b) ∠QNS
(c) ∠QOS
(d) ∠QMS
(a) The radius is perpendicular to the tangent at the point of contact.
Therefore, ∠OQP = 90°.
∠OQM = ∠OQP − ∠PQM
⇒ ∠OQM = 90° − 42°
⇒ ∠OQM = 48°
(b) By the alternate segment theorem:
∠QNM = ∠PQM = 42°
and ∠SNM = ∠PSM = 25°
Therefore, ∠QNS = ∠QNM + ∠SNM
⇒ ∠QNS = 42° + 25°
⇒ ∠QNS = 67°
(c) The angle subtended by an arc at the centre is twice the angle subtended by it at the circumference.
∠QOS = 2∠QNS
⇒ ∠QOS = 2 × 67°
⇒ ∠QOS = 134°
(d) QMSN is a cyclic quadrilateral, so its opposite angles are supplementary.
∠QMS + ∠QNS = 180°
⇒ ∠QMS + 67° = 180°
⇒ ∠QMS = 113°
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