Question

Consider the figure below:

If $\Delta BOC\cong \Delta AOD$ , and $\angle$ $DOA={30}^o$, then what is the measure of $\angle BCO$ (in degrees) ?

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Answer 1

Since, $\Delta BOC\cong \Delta AOD$, then $\angle BOC={30}^o$. From angle sum property of triangle in $\Delta BOC$, the measure of $\angle BCO$ is ${60}^o$.

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

Corresponding angles of the triangles are equal. It gives,
$\angle BOC={30}^o$
$\angle BOC+\angle BCO+\angle OBC={180}^o$ [Angle sum property] ​​​​​​​
$\angle BCO$​​​​​​​ = 180 - (90 + 30) = 180 - 120 = ${60}^o$