RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

(b) In ΔAOB
AO = BO (radii of circles)
∴∠ABO = ∠BAO = 30º
In ΔBOC

BO = CO (radii of circles)∴ ∠BCO = ∠OBC = 40º
∠ABC = ∠ABO + ∠OBC
∠ABC = 30° + 40° = 70°
2 ×∠ABC = ∠AOC ⇒x° = 140