Question

In the figure the sides $BC,CA$ and $AB$ of a $△ABC$ have been produced to $D,E$ and $F$ respectively. If $\angle ACD={105}^o$ and $\angle EAF={45}^o$, find all the angles of the $△ABC$

Answer 1

$\angle BAC=\angle EAF={45}^{\circ }(\because$ vertically opposite angles $)$

$\angle ACB={180}^{\circ }-\angle ACD={180}^{\circ }-{105}^{\circ }={75}^{\circ }$

$\angle ABC={180}^{\circ }-\angle BAC-\angle ACB={180}^{\circ }-{45}^{\circ }-{75}^{\circ }={60}^{\circ }$