Question

Given :In $\Delta ABC;AC=BC,CD$ parallel to BA and $\angle DCA={55}^o$. Calculate : $\angle ACB$ in degrees

Answer 1

Given, $AC=BC$, $CD\parallel AB$
$\angle DCA=\angle CAB={55}^{\circ }$ (Alternate angles)
Now, in $△ABC$
$AC=BC$,
Hence, $\angle CAB=\angle CBA=55$ (Isosceles triangle property)
Sum of angles = 180
$\angle ACB+\angle ABC+\angle BAC=180$
$\angle ACB+55+55=180$
$\angle ACB=180-110$
$\angle ACB={70}^{\circ }$