Question

The angles of a quadrilateral are in AP whose common difference is ${10}^{\circ }$. Find the angles.

Answer 1

It is given that the angles are in A.P. and difference between angles are ${10}^{\circ }$

$\therefore$ Let the angles be $p,p+10,p+20,p+30.$

$\Rightarrow$ $p+p+10+p+20+p+30={360}^{\circ }$ [Sum of interior angles in quadrilateral is 360${}^{\circ }$]

$\Rightarrow$ $4p+60={360}^{\circ }$

$\Rightarrow$ $4p={300}^{\circ }$

$\therefore$ $p={75}^{\circ }$

So the angles are,

$\Rightarrow$ $p={75}^{\circ }$

$\Rightarrow$ $p+10={75}^{\circ }+{10}^{\circ }={85}^{\circ }$

$\Rightarrow$ $p+20={75}^{\circ }+{20}^{\circ }={95}^{\circ }$

$\Rightarrow$ $p+30={75}^{\circ }+{30}^{\circ }={105}^{\circ }$