Answer Not Defined
In parallelogram ABCD, \(\angle\)A + \(\angle\)B = 180° ∴ \(\frac{1}{2}\)(\(\angle\)A + \(\angle\)B) = 90° That means, in triangle ABO, \(\angle\)OAB + \(\angle\)OBA = 90° ∴ The remaining angle \(\angle\)AOB = 180° − 90° = 90°
In parallelogram ABCD, \(\angle\)A + \(\angle\)B = 180° ∴ \(\frac{1}{2}\)(\(\angle\)A + \(\angle\)B) = 90° That means, in triangle ABO, \(\angle\)OAB + \(\angle\)OBA = 90° ∴ The remaining angle \(\angle\)AOB = 180° − 90° = 90°