# In Fig. 10.70, ta

As we know that the tangents drawn from an external point to a circle are equal.

Therefore, PQ = PR

Also, from the figure, PQR is an isosceles triangle, because PQ = PR

Therefore, ∠RQP =∠QRP (Because the corresponding angles of the equal sides of the isosceles triangle are equal)
And, from the angle sum property of a triangle,

∠RQP + ∠QRP + ∠RPQ = 180o

∠RQP + ∠RQP + ∠RPQ = 180o

2∠RQP +∠RPQ = 180o

2∠RQP +30o = 180o

2∠RQP = 180- 30o

2∠RQP = 150o

∠RQP = 150o/2

Therefore, ∠RQP = 75o

SR || QP and QR is a transversal

∵ ∠SRQ = ∠ PQR        …[Alternate interior angle]

∴ ∠SRQ = 75°

⇒ ∠ORP = 90°…[Tangent is Perpendicular to the radius through the point of contact]

∠ORP = ∠ORQ + ∠QRP

⇒ 90° = ∠ORQ + 75°

⇒ ∠ORQ = 15°

Similarly, ∠ RQO = 15°

In Δ QOR,

∠QOR + ∠QRO + ∠OQR = 180°

⇒ ∠QOR + 15° + 15° = 180°

⇒ ∠QOR = 150°

⇒ ∠QSR = ∠QOR/2

⇒ ∠QSR = 150°/2 = 75°

In ΔRSQ,

∠RSQ + ∠QRS + ∠RQS = 180°

⇒ 75° + 75° + ∠RQS = 180°

∠RQS = 30º

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