Question

If$\Delta BOC\cong \Delta AOD$ and $\angle$ $DOA={30}^{\circ }$, then $\angle BCO$ is equal to ___${}^{\circ }$.

Answer 1

Since, $\Delta BOC\cong \Delta AOD$,

$\angle DOA=\angle BOC={30}^{\circ }$. (corresponding angles of congruent triangles)

The sum of the angles of a triangle = 180${}^{\circ }$.

In $\Delta BOC$,

$\angle OBC={90}^{\circ }$ [Given in the diagram]

$\angle BOC={30}^{\circ }$.

$\therefore$$\angle BCO$ = 180-(90+30) = 60${}^{\circ }$.