A quadrilateral ABCD is inscribed in a circle with centre O. If ∠BOC=92 and ∠ADC=112 , then ∠ABO is equal to

Hi varun,
Opposite sides of cyclic quadrilateral when summed give $$180^{\circ \ }$$
Hence $$ABC=68^{\circ \ }\ \ AOC=2\cdot ABC=2\cdot 68=136\ $$
$$AOB+BOC+AOC=360\ \longrightarrow \ AOB=132$$
Since OA and OB are radius of triangle, $$ABO=BAO=\frac{\left(180-132\right)}{2}=24^{\circ \ }$$
Answer would be $$24^{\circ \ }$$