Question

tangent PQ and PR are drawn from external point P to a circle with Centre O such that angle are RPQ equal to 30 degree a chord is drawn parallel to the tangent PQ find angle RQS

Answer 1

HERE IS YOUR ANSWER
=>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]
Therefore, ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a triangle]

In ΔPQR ∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle]
2∠RQP + 30° = 180° 2∠RQP = 150° ∠RQP = 75°

Hence, ∠RQP = ∠QRP = 75° ∠RQP = ∠RSQ = 75° [ By Alternate Segment Theorem]
Given, RS || PQ Therefore ∠RQP = ∠SRQ = 75° [Alternate angles]
∠RSQ = ∠SRQ = 75°

Therefore 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°
Therefore, ∠RQS = 30°

Answer 2

Answer:

Step-by-step explanation:

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]

Therefore, ∠PRQ = ∠PQR [Angles opposite to equal sides are equal in a triangle]

In ΔPQR ∠RQP + ∠QRP + ∠RPQ = 180° [Angle sum property of a triangle]

2∠RQP + 30° = 180° 2∠RQP = 150° ∠RQP = 75°

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

Given, RS || PQ Therefore ∠RQP = ∠SRQ = 75° [Alternate angles]

∠RSQ = ∠SRQ = 75°

Therefore 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°

Therefore, ∠RQS = 30°