Question

The value of $x$ in degrees is equal to

Answer 1

In $△ABC$, $CA=CB$

Thus, $\angle CBA=\angle CAB={50}^o$ ...(Isosceles triangle property)
In $△ACD$,
$\angle CDA=\angle CAD=x$ ...(Isosceles triangle property)
Sum of angles = ${180}^o$
$\therefore \angle CDA+\angle CAD+\angle ADC=180$
$x+x+\angle ADC=180$
$\angle ADC=180-2x$
Also, $\angle ADC=\angle CAB+\angle CBA$ ...(Exterior angle property)
$180-2x=50+50$
$2x=80$
$x={40}^{\circ }$