In the given figure, PQ is a tangent to a circle with center O. A is the point of contact. If ∠PAB = 67°, then the measure of ∠AQB is
In the given Figure, Join OA
Now,
OA ⏊ PQ
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAP = ∠OAQ = 90° [1]
∠OAB + ∠PAB = 90°
∠OAB + 67° = 90°
∠OAB = 23°
Now,
∠BAC = 90°
[Angle in a semicircle is a right angle]
∠OAB + ∠OAC = 90°
23° + ∠OAC = 90°
∠OAC = 67°
∠OAQ = 90° [From 1]
∠OAC + ∠CAQ = 90°
67° + ∠CAQ = 90°
∠CAQ = 23° [2]
Now,
OA = OC
[radii of same circle]
∠OCA = ∠OAC
[Angles opposite to equal sides are equal]
∠OCA = 67°
∠OCA + ∠ACQ = 180° [Linear Pair]
67° + ∠ACQ = 180°
∠ACQ = 113° [3]
Now, In △ACQ By Angle Sum Property of triangle
∠ACQ + ∠CAQ + ∠AQC = 180°
113° + 23° + ∠AQC = 180° [By 2 and 3]
∠AQC = 44°