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in Circles by (30.3k points)
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PQ is a tangent to a circle with centre 0 at the point P. If A Δ OPQ is an isosceles triangle, then ∠OQP is equal to 

A. 30° 

B. 45° 

C. 60° 

D. 90°

2 Answers

+1 vote
by (30.5k points)
selected by
 
Best answer

Answer is B. 45°

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: Sum of all angles of a triangle = 180°. 

By property 1, ∆POQ is right-angled at ∠OPQ (i.e., ∠OPQ = 90°). 

∵ ∆POQ is an isosceles triangle

∴ ∠POQ = ∠OQP 

By property 2, 

∠POQ + ∠OQP + ∠QPO = 180° 

⇒ ∠POQ + ∠OQP = 180° - ∠QPO 

⇒ ∠POQ + ∠OQP = 180° - 90° 

⇒ ∠POQ + ∠OQP = 180° - 90° 

⇒ ∠POQ + ∠OQP = 90° 

⇒ ∠OQP + ∠OQP = 90° [∵∠POQ = ∠OQP] 

⇒ 2∠OQP = 90°

⇒ ∠OQP = \(\frac{90^°}{2}\) ⇒ ∠OQP = 45°

Hence, ∠OQP = 45°

+1 vote
by (105 points)
In triangle OPQ, <OPQ=90°[SINCE tangent at P]

=><POQ+<OQP=90°

=><OQP+<OQP=90°[Isosceles triangle]

=><OQP=45°

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