Question

Find $x$ in the following figure:

Answer 1

In $△ADC$,
AD = AC (given)
$\therefore \angle ADC=\angle ACD$ ($\because$ angles opposite to equal sides are equal)
$\Rightarrow \angle ADC={42}^{\circ }$
($\because \angle ACD={42}^{\circ }$(given))
$\angle ADC=\angle DAB+\angle DBA$ (Exterior angle property of a triangle)
But, $BD=DA$ (given)
$\Rightarrow \angle DAB=\angle DBA$ ($\because$ angles opposite to equal sides are equal)
$\Rightarrow \angle ADC=2\angle DBA$
$\Rightarrow 2\angle DBA={42}^{\circ }$
$\Rightarrow \angle DBA={21}^{\circ }$
Now, $x=\angle CBA+\angle BCA$ ($\because$Exterior angle property of a triangle)
$\therefore x={21}^{\circ }+{42}^{\circ }$ ($\because \angle CBA=\angle DBA={21}^{\circ }$ and $\angle BCA={42}^{\circ }$ (given))
$\therefore x={63}^{\circ }$