Question

hey guys!
please help me to solve this question... it's too urgent..
no spams please...

Answer 1

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°

Answer 2

hey here is ur ans...
as there r two tangents from a same external point. so both r equal.
so anglePQR =ang.PRQ.
so PQR =PRQ = 75°.
as RS || PQ .
PRQ = QRS = 75°
similarly,
RSQ = 75°.
so,RQS= 30°