Question

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

Answer 1

Let the angles be $(A{)}^{\circ },(A+d{)}^{\circ },(A+2d{)}^{\circ },(A+3d{)}^{\circ }$
Here, d = 10
So, $(A{)}^{\circ },(A+10{)}^{\circ },(A+20{)}^{\circ },(A+30{)}^{\circ }$ are the angles of a quadrilateral whose sum is ${360}^{\circ }$.
$\therefore (A{)}^{\circ },(A+10{)}^{\circ },(A+20{)}^{\circ },(A+30{)}^{\circ }$
$={360}^{\circ }\Rightarrow 4A=360-60A=\frac{300}{4}={75}^{\circ }$
The angles are as follows:
${75}^{\circ },(75+10{)}^{\circ },(75+20{)}^{\circ },(75+30{)}^{\circ }$,
i.e., ${75}^{\circ },{85}^{\circ },{95}^{\circ },{105}^{\circ }$