Question

In $\square$PQRS, $m\angle Q$ is ${10}^o$ less, $m\angle R$ is ${20}^o$ less and $m\angle S$ is ${30}^o$ less, than $m\angle P$, then find measure of each angle.

Answer 1

Let measure of $\angle P$ be $x$ then,

$m\angle Q=x-{10}^o,m\angle R=x-{20}^o$ and $m\angle S=x-{30}^o$

We know that sum of four angles of quadrilateral is ${360}^o.$

$\Rightarrow$ $\angle P+\angle Q+\angle R+\angle S={360}^o$

$\Rightarrow$ $x+(x-{10}^o)+(x-{20}^o)+(x-{30}^o)={360}^o$

$\Rightarrow$ $x+x-{10}^o+x-{20}^o+x-{30}^o={360}^o$

$\Rightarrow$ $4x-{60}^o={360}^o$

$\Rightarrow$ $4x={420}^o$

$\Rightarrow$ $x={105}^o$

$\Rightarrow$ $\angle P={105}^o$

$\Rightarrow$ $\angle Q=x-{10}^o={105}^o-{10}^o={95}^o$

$\Rightarrow$ $\angle R=x-{20}^o={105}^o-{20}^o={85}^o$

$\Rightarrow$ $\angle S=x-{30}^o={105}^o-{30}^o={75}^o$

$\therefore$ Measure of angles of quadrilateral are ${105}^o,{95}^o,{85}^o$ and ${75}^o.$