In the given figure,AB is the diameter of the circle with centre O. If angle DAB=70^o and angle DBC=30^o, determine angle ABD and angle CDB

Hi Satwik 
Given: ​∠DAB = 70°  and ​∠DBC = 30°
In ​ΔDAB
∠ADB = 90° {Angle in Semi circle}
∠DAB = 70° {Given}
∠ADB + ​​∠DAB + ​​∠ABD = 180° {ASP of Δ}
90° + 70° + ​∠ABD = 180°
∠ABD = 20°

In cyclic quadrilater ABCD 
∠ABC = ​∠ABD+ ​∠DBC
∠ABC = 20° + 30° = 50°
∠ABC + ​∠ADC = 180° {Opposite angles of cyclic quadrilateral are supplementary}
50° + ∠ADC = 180° 
∠ADC = 180° - 50° = 130° 

∠CDB = ∠ADC - ​∠ADB 
∠CDB = 130° - 90°  = 40° 
Therefore ​∠ABD = 20° and ​∠CDB = 40°