RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

OX is the bisector of ∠AOC
∴ ∠AOC = 2 ∠COX
OY is the bisector of ∠ BOC
∴ ∠ BOC = 2 ∠COY
∠AOC + ∠BOC = 2 ∠COY + 2 ∠COX
= 180°
∴ ∠XOY = 90
∠AOX + ∠XOY + ∠BOY = 180°
∠BOY = 180° – 90° – 20°
= 70°