Q18 of 35 Page 11

Centre of the circle is O and AP, and AQ is tangent of the circle. If ∠OPQ=20°, then what is the value of ∠PAQ?


∠OPQ = 20° …given


Radius OP is perpendicular to tangent PA at point of contact A


⇒ ∠APO = 90°


From figure ∠APO = ∠APQ + ∠OPQ


⇒ 90° = ∠APQ + 20°


⇒ ∠APQ = 70° …(a)


Consider ΔAPQ


AP and AQ are tangents to circle from A


Tangents from a point to a circle are equal


⇒ AP = AQ
hence ΔAPQ is a isosceles triangle


⇒ ∠APQ = ∠AQP


Using (a)


⇒ ∠AQP = 70°


Now


⇒ ∠APQ + ∠AQP + ∠PAQ = 180° …sum of angles of triangle


⇒ 70° + 70° + ∠PAQ = 180°


⇒ 140° + ∠PAQ = 180°


⇒ ∠PAQ = 40°


Hence ∠PAQ is 40°



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