Question

In $△ABC$, $\angle A={60}^o$, $\angle B={80}^o$ and bisectors of $\angle B$ and $\angle C$ meet at $O$. Find
(i)$\angle C$
(ii)$\angle BOC$

Answer 1

(i)$\because \angle A+\angle B+\angle C={180}^0$

$\Rightarrow {60}^0+{80}^0+\angle C={180}^0$

$\angle C={180}^0-{140}^0$

$\angle C={40}^0$

(ii) $\angle BOC={180}^0-{40}^0-{20}^0$

$={120}^0$