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Question
In figure 1, O is the centre of a circle, PQ is a chord and PT is the tangent at P.
If ∠POQ = 70°, then ∠TPQ is equal to
Options
A. 55°
B. 70°
C. 45°
D. 35°
Solution
In ΔOPQ:
OP = OQ (Radii of same circle)
⇒ ∠OQP = ∠OPQ (Equal sides have equal angles opposite to them)
∠POQ + ∠OPQ + ∠OQP = 180° (Angle sum property of triangles)
⇒ 70° + 2∠OPQ = 180°
⇒ 2∠OPQ = 180° − 70° = 110°
⇒ ∠OPQ = 55°
It is known that the tangent is perpendicular to the radius through the point of contact.
∴ ∠OPT = 90°
⇒ ∠OPQ + ∠TPQ = 90°
⇒ 55° + ∠TPQ = 90°
⇒ ∠TPQ = 90° − 55° = 35°
The correct answer is D.
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