Question

If SR is parallel to MP, <RPQ = 30°, then find <RQS ​

Answer 1

Step-by-step explanation:

It is given that,  ∠RPQ=30o 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 triangle. ]

In △PQR,

∠RQP+∠QRP+∠RPQ=180o          [Angle sum property of a triangle ]

⇒  2∠RQP+30o=180o

⇒  2∠RQP=150o

⇒  ∠RQP=75o

so ∠RQP=∠QRP=75o

⇒  ∠RQP=∠RSQ=75o           [ By Alternate Segment Theorem]

Given, RS∥PQ

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

⇒  ∠RSQ=∠SRQ=75o                                          

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

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

⇒  75o+75o+∠RQS=180o

⇒  150o+∠RQS=180o

∴    

Answer 2

Answer:

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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°

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°