The interior angles of a quadrilateral sum to 360°
Let a1 be the smallest angle and d be the common difference between the angles
So we have the following equation
129 = a1 + 3d ⇒ 129 - 3d = a1 (1)
And
a1 + (a1 + d) + (a1 + 2d ) + 129 = 360 (2)
Sub (1) into (2) and simplifying we have that
129 - 3d + (129 - 2d) + ( 129 - d) + 129 = 360
516 - 6d = 360 rearrrange as
516 - 360 = 6d
156 = 6d divide both sides by 6
26 = d
And using (1) we have that 129 - 3(26) = a1 = 51°
So.....the second largest angle is 51 + 2(26) = 103°