Question

tangents PQ and PR are drawn from external point Pramodani to a circle with centre O, such that angle angleRPQ=30°. A chord RAHUL is drawn parallel to the tangents PA. find AngleRQS​

Answer 1

hey mate!

Given ∠RPQ=30° and PR and PQ are tangents drawn from P to the same circle.

Hence PR = PQ [Since tangents drawn from an external point to a circle are equal in length]

∴ ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a Δ]

In ΔPQR

∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a Δ]

2∠RQP + 30° = 180°

2∠RQP = 150°

∠RQP = 75°

so ∠RQP = ∠QRP = 75°

∠RQP = ∠RSQ = 75°  [ By Alternate Segment Theorem]

Given, RS || PQ

∴ ∠RQP = ∠SRQ = 75°    [Alternate angles]

∠RSQ = ∠SRQ = 75°                                          

∴ QRS is also an isosceles triangle.     [Since sides opposite to equal angles of a triangle are equal.]

∠RSQ + ∠SRQ + ∠RQS = 180° [Angle sum property of a triangle]

75° + 75° + ∠RQS = 180°

150° + ∠RQS = 180°

∴ ∠RQS = 30°

Answer 2

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