Question

$BO$ and $CO$ are bisectors of Exterior Angle $B$ and $C$ respectively $\angle BAC={70}^{\circ }\angle ABC={40}^{\circ }$ find $\angle BOC$

Answer 1

$2\angle XBA+{40}^o={180}^o\to \angle XBA={70}^o$ (Linear pair)

$\angle ACB={180}^o-{70}^o-{40}^o={70}^o$ (sum of angles $={180}^o$)

$2\angle YCA+{70}^o={180}^o\to \angle YCA={55}^o$ (Linear pair)

$\angle CBO+{40}^o+{70}^o={180}^o\to \angle CBO={70}^o$ (Linear pair)

$\angle BCO+{70}^o+{55}^o={180}^o\to \angle BCO={55}^o$ (Linear pair)

$\angle CBO+\angle BCO+\angle BOC={180}^o$ (sum of angles in $△={180}^o$)

$\angle BOC={180}^o-{70}^o-{55}^o={55}^o$