Question

Find the value of $x^{\circ }$ in the following figure:

Answer 1

$\angle OAC+\angle OCA={180}^{\circ }-{100}^{\circ }$
$={80}^{\circ }$
$\angle OAC=\frac{1}{2}\times {80}^{\circ }\dots [$ Since $OA=O{C}_r(\therefore \angle OAC=\angle OCA)]$
$={40}^{\circ }$
$\angle OBA+\angle OAB={180}^{\circ }-{140}^{\circ }$
$={40}^{\circ }\dots [$ Since $\angle OBA=\angle OAB$, since $OB=OA]$
$\therefore \angle OAB=\frac{{40}^{\circ }}{2}$
$={20}^{\circ }$
$\angle BAC=\angle OAB+\angle OAC$
$={20}^{\circ }+{40}^{\circ }$
$x={60}^{\circ }$