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In the given figure, PQR is a tangent to the circle at Q, whose center is O and AB is a chord parallel to PR such that ∠BQR = 70°. Then, ∠AQB =?
In given figure, as PR is a tangent
OQ ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
⟹LQ ⏊ PR
⟹LQ ⏊ AB
[As, AB || PR]
AL = LB
[Perpendicular from center to the chord bisects the chord]
Now,
∠LQR = 90°
∠LQB + ∠BQR = 90°
∠LQB + 70° = 90°
∠LQB = 20°…[1]
In △AQL and △BQL
∠ALQ = ∠BLQ [Both 90° as LQ ⏊ AB]
AL = LB [Proved above]
QL = QL [Common]
△AQL ≅ △BQL
[Side - Angle - Side Criterion]
∠LQA = ∠LQB
[Corresponding parts of congruent triangles are congruent]
∠AQB = ∠LQA + ∠LQB = ∠LQB + ∠LQB
= 2∠LQB = 2(20) = 40° [By 1]