Question

Solve this question please fast. Tomorrow I'm having test.So, please.....

Answer 1

For diagram pls refer to attachment




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°
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Answer 2

PQ=QR (Since tangents drawn from ext. pt. to circle is equal)
PQR is isosceles ∆.
So, RQP = QRP
So, 2 RQP + RPQ = 180°
=>RQP =QRP=75°
RQP =RSQ =75° (Angle in alternate segment theorem)
RQP = SRQ =75° (Alternate Angles)
So, 75°+75°+RQS = 180°
=> RQS = 30°