Question 911466
In quadrilateral ABCD, m/A=(x-10), m/B=(2x+10), m/C=(2x-70) m/D(3x-50)

Sum of angles of quadrilateral = 360 deg

(x-10)+(2x+10)+(2x-70)+(3x-50)=360

8x-120=360
6x=360+120
8x=480

x=60

 m/A=(x-10),=50
 m/B=(2x+10),130
 m/C=(2x-70) 50
m/D(3x-50)130

The opposite angles are equal
Hence its a parallelogram