Please solve these folliwing questions

We know that,  the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

⇒ ∠ TQP = ∠ TPQ ...(1) ( In a triangle, equal sides have equal angles opposite to them )

∠ TQP + ∠ TPQ + ∠ PTQ = 180º ( Angle sum property )

∴ 2 ∠ TPQ + ∠ PTQ = 180º (Using(1))

⇒ ∠ PTQ = 180º – 2 ∠ TPQ ...(1)

We know that,  a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠ OPT = 90º

⇒ ∠ OPQ + ∠ TPQ = 90º

⇒ ∠ OPQ = 90º – ∠ TPQ

⇒ 2 ∠ OPQ = 2(90º – ∠ TPQ) = 180º – 2 ∠ TPQ ...(2)

From (1) and (2), we get

∠ PTQ = 2 ∠ OPQ
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