Question

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

• A. ${94}^o$
• B. ${54}^o$
• C. ${40}^o$
• D. ${44}^o$

Answer 1

The correct option is C ${40}^o$

Given that,

the base$=BC$ such that,

$\angle ACD={126}^o$ and $\angle ABE={94}^o$

We know that,

$\angle ABE+\angle ABC={180}^o$

$\angle ABC={180}^o-\angle ABE$ $={180}^o-{94}^o$

$\angle ABC={86}^o$

Again, we know that

$\angle ACB+\angle ACB={180}^o$

$\angle ACB={180}^o-\angle ACD$ $={180}^o-{126}^o$

$\angle ACB={54}^o$

In $\Delta$ABC

$\angle ABC+\angle ACB+\angle BAC={180}^o$

${86}^o+{54}^o+\angle BAC={180}^o$

$\angle BAC={180}^o-{140}^o$

$\angle BAC={40}^o$ Hence, this is the answer.