Question

In $\Delta ABC,AB=AC,\angle A={40}^{\circ },O$ is a point inside $\Delta ABC$ such that $\angle OBC=\angle OCA$. Find the measure of $\angle BOC.$

• A. ${110}^{\circ }$
• B. ${35}^{\circ }$
• C. ${140}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

The correct option is A ${110}^{\circ }$

Given: AB = AC
$\therefore \angle ABC=\angle ACB$ ...(i) {Angle opposite equal sides}
Also, $\angle OBC=\angle OCA$ ...(ii) {given}
Subtract (i) from (ii)
$\angle OBA=\angle OCB$ ...(iii)
Let $\angle OBC=\angle OCA=x$
and $\angle OBA=\angle OCB=y$
$\angle ABC=\angle ACB=x+y$
In $△ABC$,
$\angle ABC+\angle BAC+\angle BCA={180}^{\circ }$
{Using Angle Sum Property}
40${}^{\circ }$ + (x + y) + (x + y) =180${}^{\circ }$
$\Rightarrow 2(x+y)={140}^{\circ }$
$\Rightarrow x+y={70}^{\circ }$
In $△OBC$,
$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$
$\Rightarrow x+y+\angle BOC={180}^{\circ }$
$\Rightarrow \angle BOC={180}^{\circ }-(x+y)$
$\Rightarrow \angle BOC={180}^{\circ }-{70}^{\circ }={110}^{\circ }$