Question

In square $ABCD$, $\angle A={120}^o$. If $\angle B=\angle C=\angle D$, then find the measures of $\angle B$.

Answer 1

We know, by angle sum property, the sum of angles of a quadrilateral is ${360}^o$.

The given $\angle A={120}^o$ and let $\angle B=\angle C=\angle D=x$.

Then, $\angle A+\angle B+\angle C+\angle D={360}^o$

$⟹$ ${120}^o+x+x+x={360}^o$

$⟹$ ${120}^o+3x={360}^o$

$⟹$ $3x={360}^o-{120}^o$

$⟹$ $3x={240}^o$

$⟹$ $x=\frac{{240}^o}{3}={80}^o$.

Therefore, $\angle B=x={80}^o$.