In the given figure, EF = CE = CA, What is the value (in degrees) of $$\angle$$EAC?

Given : EF = CE = CA

=> $$\angle$$ CAE = $$\angle$$ CEA = $$x$$ and $$\angle$$ ECF = $$\angle$$ EFC = $$y$$

To find : $$\angle$$ EAC = $$x=?$$

Solution : Using exterior angle property, => $$\angle$$ CAE + $$\angle$$ CFE = $$\angle$$ ACD

=> $$x+y=96^\circ$$ -----------------(i)

Also, $$\angle$$ CEF = $$(180^\circ-2y)=180^\circ-x$$

=> $$x=2y$$ ------------(ii)

Substituting it in equation (i), => $$2y+y=3y=96^\circ$$

=> $$y=\frac{96}{3}=32^\circ$$

$$\therefore$$ $$x=2\times32=64^\circ$$

=> Ans - (B)