Question

In the adjoining figure, find the value of x.

Answer 1

In $\Delta ADC$
$\angle CAD+\angle ADC+\angle DCA={180}^{\circ }$ [Angle sum property]
$\Rightarrow \angle CAD+\angle ADC+{64}^{\circ }={180}^{\circ }$
$\Rightarrow \angle CAD+\angle ADC=({180}^{\circ }-{64}^{\circ })={116}^{\circ }$
But CD = CA $\Rightarrow \angle CAD=\angle ADC$ [ $\because$ $\angle s$ opposite to equal sides of a $\Delta$ are equal]
So, $\angle CAD=\angle ADC=\frac{{116}^{\circ }}{2}={58}^{\circ }$
Now, $\angle ADC=\angle ABD+\angle DAB$
[Ext. $\angle$ of a $\Delta$ = sum of int. opp. $\angle s$]
But, $AD=BD\Rightarrow \angle ABD=\angle DAB$.
So, $\angle ADC=2\angle DAB$
$\Rightarrow \angle DAB=\frac{1}{2}\angle ADC$
$\Rightarrow x=\frac{1}{2}\times {58}^{\circ }={29}^{\circ }$
Hence, $x={29}^{\circ }$