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
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(a) From figure,
⇒ ∠OQP = 90° (Tangent is perpendicular to radius at the point of contact)
⇒ ∠OQM = ∠OQP - ∠PQM
⇒ ∠OQM = 90° - 42° = 48°.
Hence, ∠OQM = 48°.
(b) From figure,
⇒ ∠QNM = ∠PQM = 42° (By alternate segment theorem)
⇒ ∠SNM = ∠PSM = 25° (By alternate segment theorem)
⇒ ∠QNS = ∠QNM + ∠SNM
⇒ ∠QNS = 42° + 25° = 67°.
Hence, ∠QNS = 67°.
(c) We know that,
Angle subtended by an arc at the center is twice the angle subtended by the arc at any other point of the circle.
⇒ ∠QOS = 2∠QNS
⇒ ∠QOS = 2 × 67° = 134°.
Hence, ∠QOS = 134°.
(d) From figure,
QMSN is a cyclic quadrilateral.
We know that,
Sum of opposite angles of a cyclic quadrilateral is 180°.
⇒ ∠QMS + ∠QNS = 180°
⇒ ∠QMS + 67° = 180°
⇒ ∠QMS = 180° - 67° = 113°.
Hence, ∠QMS = 113°.