Question

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?

Answer 1

Answer:

∠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°