Question

In the figure given below, $\angle BCD={60}^{\circ }$. Enter the measure of $\angle BDA$ in degrees?

• A. 30

Answer 1

The correct option is A 30
In $△$ABD and $△$CDB,
DA = BC (given)
BD = DB (common side)
$\angle ABD=\angle CDB={90}^{\circ }$

Hence, $△ABD\cong △CDB$
(by RHS criteria for congruence of triangles)

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

By ange sum property of triangle,
$\angle BDA+\angle ABD+\angle DAB={180}^{\circ }$
$\angle BDA={180}^{\circ }-\angle ABD+\angle DAB$
$⟹\angle BDA={180}^{\circ }-({90}^{\circ }+{60}^{\circ })$
$⟹\angle BDA={180}^{\circ }-{150}^{\circ }$
$⟹\angle BDA={30}^{\circ }$