RRB Non Technical Model Papers 8 with solutions

∠BOC + ∠AOB + ∠COA = 360°
(angles around a point) In Δ BOC, ΔAOB and Δ COA OB = OC = OA(radius of circle) BC = AB = CA(sides of equilateral triangle) ∴ ΔBOC, Δ AOB and Δ COA are congruent.∴ ∠BOC =∠AOB =∠COA ∴ ∠BOC = ∠AOB = ∠COA= 120°