In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70^∘ , then ∠TPQ is equal to [CBSE 2011]
(a) 35^∘
(b) 45^∘
(c) 55^∘
(d) 70^∘

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Solution

We know that the radius and tangent are perperpendular at their point of contact
Since, OP = OQ
∵POQ is a isosceles right triangle
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ [Angle sum property of a triangle]
⇒ 70^∘ + 2∠OPQ = 180^∘
⇒ ∠OPQ = 55^∘
Now, ∠TPQ + ∠OPQ = 90^∘
⇒ ∠TPQ = 35^∘
Hence, the correct answer is option (a).

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