Question

In a triangle PQR,$\angle P={60}^{\circ }$ and $\angle Q=\angle R$, find $\angle R$.

Answer 1

Consider $\angle Q=\angle R=x$

$\angle P={60}^{\circ }$

We can write it as

$\angle P+\angle Q+\angle R={180}^{\circ }$

Substituting the values

${60}^{\circ }+x+x={180}^{\circ }$

By further calculation

${60}^{\circ }+2x={180}^{\circ }$

$2x={180}^{\circ }–{60}^{\circ }={120}^{\circ }$

So we get

x$={120}^{\circ }/2={60}^{\circ }$

$\angle Q=\angle R={60}^{\circ }$

Therefore, $\angle R={60}^{\circ }$