Question

In the given figure, TP and TQ are two tangents to the circle with centre O, touching at A and C respectively. If ∠BCQ = 55° and ∠BAP = 60°, find:

1. ∠OBA and ∠OBC 
2. ∠AOС 
3. ∠ATС

Answer 1

Given:

TP and TQ are tangents touching the circle at points A and C, respectively.

∠BCQ = 55°

∠BAP = 60°

(i) Find ∠OBA and ∠OBC

Tangent is perpendicular to the radius at the point of contact. So, OC ⊥, TC, and OA ⊥ TA. Hence, ∠QCO = 90° and ∠PAO = 90°

Calculate ∠BCO:

∠BCO = ∠QCO − ∠BCQ = 90° − 55° = 35°

Triangle BOC is isosceles (OB = OC radii). So,

∠OBC = ∠BCO = 35°

Calculate ∠BAO:

∠BAO = ∠PAO − ∠BAP = 90° − 60° = 30°

Triangle AOB is isosceles (OA = OB radii). So,

∠OBA = ∠BAO = 30°

(ii) Find ∠AOC

∠ABC = ∠OBA + ∠OBC

= 30° + 35° = 65°

Angle at the center is double the angle at the circumference: 

∠AOC = 2 × ∠ABC

= 2 × 65° = 130°

(iii) Find ∠ATC

Quadrilateral AOCT sum of angles = 360°

∠AOC + ∠OCT + ∠ATC + ∠OAT = 360°

OC ⊥ TC and OA ⊥ TA give:

∠OCT = ∠OAT = 90°

Substitute and solve:

130° + 90° + ∠ATC + 90°

= 360° ∠ATC

= 360° − 310°

= 50°