Q. 174.4( 7 Votes )
In Fig., tangents
tangents PQ and PR are drawn from an external point P.
To find: ∠RSQ
1.) The length of two tangents drawn from an external point are equal.
2.) Tangent to a circle at a point is perpendicular to the radius through the point of contact.
As P is external point and PR and PQ are tangents,
By theorem (1) stated above,
PQ = PR
As angles opposite to equal sides are equal.
⇒ ∠RQP = ∠QRP
So PQR is an isosceles triangle.
By angle sum property,
⇒ 2∠RQP+30° =180°
From the theorem (2) stated above,
∠OQP = ∠ORP = 90°
We know sum of angles of a quadrilateral is equal to 360°.
∠OQP + ∠ORP + ∠RPQ + ∠QOR = 360°
⇒ 90° + 90° + 30° + ∠QOR = 360°
⇒ 210° + ∠QOR = 360°
⇒ ∠QOR = 150°
As we know angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc,
Also ∠QSR = ∠SQU (alternate angles)
∠SQU = 75°
Angle on a straight line is 180°,
∠SQU + ∠RQS + ∠PQR = 180°
⇒ 75° + ∠RQS + 75° = 180°
⇒ 150° + ∠RQS = 180°
⇒ ∠RQS = 30°
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