Question

In $\Delta$PQR, $\angle P={70}^o$, $\angle Q={65}^o$, then find $\angle R$.

Answer 1

Sum of internal angles of triangle is ${180}^o$

$\therefore$ In $△PQR$

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

$\therefore {70}^o+{65}^0+\angle R={180}^o$

$\therefore \angle R={45}^o$