Question

Triangle PAB is formed by three tangents to circle O and ∠APB = 40°. Then ∠AOB equals 

(a) 45° 

(b) 60° 

(c) 70° 

(d) 55°

Answer 1

Answer : (c) 70º 

∠P = 40° 

∴ ∠PAB + ∠PBA = 180° – 40° = 140°

∠TAS = 180° – ∠PAB 

∠RBS = 180° – ∠PBA 

∴ ∠TAS + ∠RBS = 360° – (∠PAB + ∠PBA) 

= 360° – 140° = 220° 

Since OA and OB bisect angles TAS and RBS respectively. 

∠OAS + ∠OBS = 1 2 × 220° = 110° 

∴ ∠AOB = 180° – 110° = 70°.