Question

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

Answer 1

Answer:

∠RQS = 30°

Step-by-step explanation:

PQ = PR

Since tangents drawn from an external point to a circle are equal.

And PQR is an isosceles triangle

thus, ∠RQP = ∠QRP

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

2∠RQP + 30° = 180°

2∠RQP = 150°

∠RQP = ∠QRP = 75°

∠RQP = ∠RSQ = 75°  [ Angles in alternate Segment Theorem states that angle between chord and tangent is  equal to the angle in the alternate segment]

RS is parallel to PQ

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

∠RSQ = ∠SRQ = 75°

Therefore QRS is also an isosceles triangle

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

75° + 75° + ∠RQS = 180°

150° + ∠RQS = 180°

∠RQS = 30°

Answer 2

Answer:

Hope this helps you

Step-by-step explanation:

It is given that, ∠RPQ=30 degree

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,

In △PQR,∠RQP+∠QRP+∠RPQ=180 degree [Angle sum property of a triangle ]

2∠RQP+30 degree =180degree

⇒ 2∠RQP=150degree

⇒ ∠RQP=75 degree

so ∠RQP=∠QRP=75 degree

⇒ ∠RQP=∠RSQ=75 degree [ By Alternate segment theorem]

Given, RS∥PQ

Given, RS∥PQ∴ ∠RQP=∠SRQ=75 degree [Alternate angles]

[Alternate angles]⇒ ∠RSQ=∠SRQ=75 degree

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

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

⇒ 75 +75 +∠RQS =180 degree

⇒ 150 degree +∠RQS=180 degree

∴ ∠RQS=30 degree