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? (2 marks)

Answer 1

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 lie on a straight line) [1 mark]

$\angle$BOC = 180${}^{\circ }$ - [$\angle$COD + $\angle$AOD] [0.5 marks]

$\Rightarrow$ x${}^{\circ }$ = 180${}^{\circ }$ – (40${}^{\circ }$ + 80${}^{\circ }$)
$\Rightarrow$ x${}^{\circ }$ = 180${}^{\circ }$– 120${}^{\circ }$
$\Rightarrow$ x${}^{\circ }$ = 60${}^{\circ }$ [0.5 marks]