Question

A ray OC & OD is constructed on the line segment AB such that ∠AOD = 40${}^{\circ }$ and ∠COD = 80${}^{\circ }$ and ∠BOC = x${}^{\circ }$ . What will ∠BOC be?

• A. x^o
• B. 180^o - x^o
• C. 90^o - x^o
• D. 360^o - x^o

Answer 1

The correct option is B

180^o - x^o

AB is a straight line and the ray OC and OD is drawn on the line segment AB. Therefore,

$\angle$AOD + $\angle$COD + $\angle$BOC= 180${}^{\circ }$ ($\angle$AOD,$\angle$COD and $\angle$BOC are linear pair of angles)

$\angle$BOC = 180${}^{\circ }$ - [$\angle$COD + $\angle$AOD]

x${}^{\circ }$ = 180${}^{\circ }$ – [40${}^{\circ }$ + 80${}^{\circ }$]
x${}^{\circ }$ = 180${}^{\circ }$ – [120${}^{\circ }$]
x${}^{\circ }$ = 60${}^{\circ }$