Question

In the figure given below, $\angle BCD={60}^{\circ }$,
then the measure of $\angle$BDA is.

Answer 1

In triangles, $△$ABD and $△$CDB,
DA = BC (given)
BD = DB (common side)
$\angle ABD=\angle CDB={90}^{\circ }$
Hence,
$△ABD\cong △CDB$
by RHS criterion for congruency of triangles.

So,
$\angle$DAB = $\angle$BCD=60${}^{\circ }$

Sum of all the angles of a triangle is
${180}^{\circ }$

$\angle BDA={180}^{\circ }-\angle ABD+\angle DAB$
$⟹\angle BDA={180}^{\circ }-({90}^{\circ }+{60}^{\circ })$
$⟹\angle BDA={180}^{\circ }-{150}^{\circ }$
$⟹\angle BDA={30}^{\circ }$