Q3 of 14 Page 1

In figure 1, O is the centre of a circle, PQ is chord and PT is the tangent at P. If ∠POQ = 70ᵒ, then ∠TPQ is equal to


By angle sum property of a triangle, we have


∠POQ + ∠OQP + ∠OPQ = 180ᵒ (i)


Also, OQ = OP [Radii of same circle]


⇒ ∠OQP = ∠OPQ


From equation (i)


70° + ∠OQP + ∠OQP = 180°


⇒ ∠OQP = 55°


Also, we know tangent drawn at a point on the circle is perpendicular to the radius through point of contact.


⇒ ∠TPO = 90ᵒ


Also, ∠TPO = ∠TPQ + ∠QPO


⇒ 90° = ∠TPQ + 55°


⇒ ∠TPQ = 90ᵒ – 55ᵒ = 35ᵒ


Hence, option D is correct

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