#### 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°

Answer(A) $100^{\circ}$
Solution
Given :$\angle QPR= 50^{\circ}$
We know that tangent is perpendicular to radius.
Hence $PR\perp PO$
$\angle$OPR = $90^{\circ}$
$\angle$OPR = $\angle$OPQ + $\angle$QPR
$90^{\circ}$ = $\angle$ OPQ + $50^{\circ}$
$\angle$OPQ = $40^{\circ}$
$\angle$OPQ = $\angle$OQD         $\left ( \because \, OP= OQ \, radius\, of\, circle\right )$
Hence , $\angle$OPQ = $40^{\circ}$
We know that the sum of interior angles of a triangle is $180^{\circ}$
In $\bigtriangleup OPQ$
$\angle$O+$\angle$Q +$\angle$P = $180^{\circ}$
$\angle$O +40+40 = 180
$\angle$o = $100^{\circ}$
Hence $\angle$OPQ = $100^{\circ}$