Question

PQ is a common chord of two circles. APB is a secant line joining points A and B on the two circles. Two tangents AC and BC are drawn. If ∠ACB = 45°, then what is ∠AQB equal to?

(a) 75° 

(b) 90° 

(c) 120° 

(d) 135° 

Answer 1

Answer : (d) 135º 

Since the tangents drawn on the two given circles, from the same external point are equal, CA = CB

⇒ ∠CAB = ∠CBA = x (say) 

In ∆ CAB, 45° + x + x = 180°

⇒ 2x = 135° 

⇒ x = 67\(\frac{1}{2}\)º 

∠AQP = ∠BQP 

= x =   67\(\frac{1}{2}\)º  (Alternate Segment Theorem) 

∠AQB = ∠AQP + ∠BQP =    67\(\frac{1}{2}\)º  +   67\(\frac{1}{2}\)º   = 135º