Question

tangent PQ and PR are drawn from an external point P to a circle with Centre O and angle rpq = 20 degree . A chord RS is drawn parallel to tangent PQ . find angle RQS

Answer 1

Tangents drawn from an external point to a circle are equal

Hence, PQ = PR

And PQR is an isosceles triangle

ThereforeRQP =QRP

RQP +QRP +RPQ = 180[Angles in a triangle]

2RQP + 30 = 180°

2RQP = 150°

RQP =QRP = 75°

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

RS II PQ

Therefore RQP = SRQ= 75° [They are alternate angles]

RSQ = SRQ = 75°

Therefore QRS is also an isosceles triangle

RSQ + SRQ + RQS = 180° [Angles in a triangle]

75° + 75° + RQS = 180°

150 °+ RQS = 180°

RQS = 30°

Required Answer: RQS = 30°

Answer 2

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

So, PQ = PR, and PQR is an isosceles triangle.

So, ∠RQP = ∠QRP

Again, ∠RQP + ∠QRP + ∠RPQ = 180

=> 2∠RQP + 30 = 180

=> 2∠RQP = 180 - 30

=> 2∠RQP = 150

=> ∠RQP = 150/2

=> ∠RQP = 75

=> ∠RQP = ∠QRP = 75

and ∠RQP = ∠RSQ = 75

=> ∠RQP = ∠SRQ = 75 {alternalte angles}

So, QRS is an isosceles triangle. {Since sides opposite to equal angles of a triangle are equal.}

Now, ∠RSQ + ∠SRQ + ∠RQS = 180° {Angle sum property of a triangle}

=> 75 + 75 + ∠RQS = 180

=> 150 + ∠RQS = 180

=> ∠RQS = 180 - 150

=> ∠RQS = 30°