Answer:
Given,
d = 10
Sum of all angles in a quadrilateral = 360
Assume the angles are a – 3d, a – d, a + d, a + 3d
a – 2d + a – d + a + d + a + 2d = 360
4a = 360
a = 90… (i)
(a – d) – (a – 3d) = 10
2d = 10
d = 10/2
= 5
Hence, the angles are a – 3d, a – d, a + d, a + 3d which is 75o, 85o, 95o, 105o