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In Figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to


(A) 100°          (B) 80°              (C) 90°             (D) 75°

Answers (1)

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Answer(A) 100^{\circ}
Solution
Given :\angle QPR= 50^{\circ} 
We know that tangent is perpendicular to radius.
Hence PR\perp PO
\angleOPR = 90^{\circ}
\angleOPR = \angleOPQ + \angleQPR
90^{\circ} = \angle OPQ + 50^{\circ}
\angleOPQ = 40^{\circ}
\angleOPQ = \angleOQD         \left ( \because \, OP= OQ \, radius\, of\, circle\right )
Hence , \angleOPQ = 40^{\circ}
We know that the sum of interior angles of a triangle is 180^{\circ}
In \bigtriangleup OPQ
\angleO+\angleQ +\angleP = 180^{\circ}
\angleO +40+40 = 180
\angleo = 100^{\circ}
Hence \angleOPQ = 100^{\circ}

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