Question

In the figure, if O is circumcentre of $\Delta ABC$ then find the value of $\angle OBC+\angle BAC$.

Answer 1

In the figure, join OC

$\because$ O is the circumcentre of $\Delta ABC$

$\therefore OA=OB=OC$

$\because \angle CAO={60}^o$

Now, $\angle BOC=\angle BOA+\angle AOC$

$={80}^o+{60}^o={140}^o$

and in $\Delta OBC$, OB = OC

$\angle OCB=\angle OBC$

But $\angle OCB+\angle OBC={180}^o-\angle BOC={180}^o-{140}^o={40}^o$

$\Rightarrow \angle OBC+\angle OBC={40}^o$

$\therefore \angle OBC=\frac{{40}^o}{2}={20}^o$

$\angle BAC=OAB+\angle OAC$

$={50}^o+{60}^o={110}^o$

$\therefore \angle OBC+\angle BAC={20}^o+{110}^o={130}^o$