Question

One of the angles of a triangle has measure ${80}^o$ and the other two angles are equal.

Answer 1

We have,

In $\Delta$ABC,

$\angle A={80}^o$

$\angle B=\angle C=x$

By angle sum

$\angle A+\angle B+\angle C={180}^o$

$80+x+x={180}^o$

$2x+80=180$

$2x=180-80$

$2x=100$

$x=\frac{100}{2}$

$x=50$

$\angle B=\angle C=x={50}^o$

Hence, this is answer.