Question

The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are ${94}^{\circ }$ and ${126}^{\circ },$ Then , $\angle BAC=?$

Answer 1

In $\Delta ABC,$ base BC is produced both ways and $\angle ACD={94}^{\circ },\angle ABE={126}^{\circ }$

Ext. $\angle ACD=\angle BAC+\angle ABC$

$\Rightarrow {94}^{\circ }=\angle BAC+\angle ABC$

Similarly, $\angle ABE=\angle BAC+\angle ACB$

$\Rightarrow {126}^{\circ }=\angle BAC+\angle ACB$

Adding,

${94}^{\circ }+{126}^{\circ }=\angle BAC+\angle ABC+\angle ACB+\angle BAC$

${220}^{\circ }={180}^{\circ }+\angle BAC$

$\therefore \angle BAC={220}^{\circ }-{180}^{\circ }={40}^{\circ }$