Math, asked by Anonymous, 4 months ago

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

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Answers

Answered by varchaswjaiswal7299

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

∴

Answered by Anonymous

Answer:

$\huge{\underline{\underline{\boxed{\sf{\purple{Answer✓}}}}}}$ $\sqrt[3]{ - 64 \times 125}$

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.