Given,

ABCD is a parallelogram

Diagonals of parallelogram intersects at O

∠BOC = 90° , ∠BDC = 50°

∵ ∠BOC+∠AOB = 180° [Linear pair of angles]

90°+∠AOB = 180°

=∠AOB = 180°-90° = 90°

∠BDC = ∠OBA = 50° [Alternate angles]

In ∆AOB

∠AOB+∠OBA+∠OAB = 180°

= 90°+50°+∠OAB = 180°

= ∠OAB = 180°-140° = 40°