the four angles of a quadrilateral form an a.p.if the sum of the first three angles is twice the fourth angle,then find all the angles.

Let the angles be ∠A=a, ∠B=a+d, ∠C=a+2d, ∠D=a+3d
∠A+∠B+∠C+∠D=360°[angle sum property of quadrilateral]
a+a+d+a+2d+a+3d=360°
4a+6d=360  ...(i)
 

 

Given that the sum of the first three angles is twice the fourth angle.

∠A+∠B+∠C=2∠D
a+a+d+a+2d=2(a+3d)
3a+3d=2a+6d
a=3d ...(ii)

 

So, from (i) and (ii), we have

4(3d)+6d=360

12d+6d=360

18d=360

d=20

So, a=3d=3(20)=60

∠A=a=60°
∠B=a+d=60+20=80°
∠C=a+2d=60+2(20)=100°
∠D=a+3d=60+3(20)=120°