Question

In parallelogram PQRS, $\angle$Q$=(4x-5{)}^o$ and $\angle$S$=(3x+10{)}^o$. Calculate: $\angle$Q and $\angle$R.

Answer 1

We know that, opposite angles of a parallelogram are equal.

$\therefore$ $\angle Q=\angle S$

$\Rightarrow$ $4x-5=3x+10$

$\Rightarrow$ $4x-3x=10+5$

$\Rightarrow$ $x=15$

$\Rightarrow$ $\angle Q=(4x-5{)}^o=(4\times 15-5{)}^o={55}^o$

We know that, sum of adjacent angles of parallelogram is ${180}^o$

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

$\Rightarrow$ ${55}^o+\angle R={180}^o$

$\therefore$ $\angle R={125}^o$