Question

Find the measure of an exterior angle at the base of an isosceles triangle whose vertex angle measures ${50}^o$.

• A. ${100}^o$
• B. ${110}^o$
• C. ${115}^o$
• D. ${65}^o$

Answer 1

The correct option is C ${115}^o$

The $2$ base angles of an isosceles triangle are equal, so we'll represent each as $x$.

Now, by using the angle sum property,

$x+x+{50}^o={180}^o$

$\Rightarrow 2x={180}^o-{50}^o$

$\Rightarrow 2x={130}^o$

$\Rightarrow x={65}^o$

So, exterior angle is equal to the sum of the two non-adjacent interior angles.

Therefore,

$y=x+{50}^o$

$y={50}^o+{65}^o$

$={115}^o$

Hence, option $C$ is correct.