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: $AB=AC$
$\angle ABC=\angle ACB\dots \dots \left(i\right)$ (Angle opposite equal sides)
Also, $\angle OBC=\angle OCA$
Let $\angle OBC=x=\angle OCA$ and $\angle OBA=y$

Since , From (i)
$\angle ABC=\angle ACB$
Therefore, $\angle ABC=x+y=\angle ACB$

In $\Delta ABC$, $\angle ABC+\angle BAC+\angle BCA={180}^{\circ }$ (Using angle sum property)
$\Rightarrow {40}^{\circ }+(x+y)+(x+y)={180}^{\circ }$
$\Rightarrow 2(x+y)={140}^{\circ }$
$\Rightarrow (x+y)={70}^{\circ }$

In $\Delta OBC$, $\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$
$x+y+\angle BOC={180}^{\circ }$
$\angle BOC={180}^{\circ }-(x+y)$
$\Rightarrow \angle BOC={180}^{\circ }-{70}^{\circ }={110}^{\circ }$