Question

find the sizes of the angles of a parallelogram if one angle is 20 less than twice the smallest angle

Answer 1

let the the smallest angle be x 
than the other angle = 2x - 20
as we know in parallelogram the angle opposite sides are equal we can say that the angles of the parallelogram are=
x,2 x-20,x ,2 x-20
the sum of all the angles are 360 in a parallelogram
so, x+2 x-20+x+2 x-20=360
6x -20 = 360
6x=360+40
6 x = 400
x =400/6
x = 66 2/3 (when we simplify 4/6)
2 x -20 = (66 2/3 ×2) - 20
2x - 20 = 113 1/3
∴ the angles are 66 2/3 ,113 1/3 , 66 2/3 ,113 1/3 ( ans)

Answer 2

Let the smallest Angle be x

Other angle be = 2x-20

In parallelogram opposite angles are equal.

Sum of the consecutive interior angles = 180°

X+ (2x-20)= 180°

3x-20°=180°

3x=180+20

3x=200

X= 200/3
X= 66.666......= 66.7°(approx.)

Smallest angle= 66.7°

Other angle = 2x-20
= 2×66.7-20= 133.4-20= 113.4°

Other angle is 113.4°

The 4 angles of a parallelogram are 66.7° ,113.4°, 66.7° , 113.4°