SSC CGL 2012 Tier 1 1 July NZ Morning III

Question 104

A, 0, B are three points on a line segment and C is a point not lying on AOB. If ∠AOC = 40° and OX, OY are the internal and external bisectors of ∠AOC respectively, then ∠BOY is

Solution

OX is the bisector of ∠AOC = 2 ∠COX.
∴ ∠BOC = 2 ∠COY
∴ ∠AOC + ∠BOC
2∠COX + 2∠COY = 180°
2(∠COX + ∠COY) = 180°
2∠XOY = 180°
∠XOY = 90°
∴ ∠AOX + ∠XOY + ∠BOY = 180°
∠BOY = 180° - 20° - 90° = 70°.


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