Question

Let's see who can answer this.In the given figure AB is the diameter of the circle with Centre O. DO is parallel to CB and angle DCB is equal to 120 degree find angle DAB, angle DBA ,angle DBC and Angle ADC

Answer 1

angle DAB will be 60 degree ....... opposite angles of cyclic quad.
angle ADB will be 90 .......angle in semicircle
Angle DBA will be 30 ....... remaining angle of triangle
Angle ODB = Angle OBD = 30 ---- isosceles triangle OBD
Now DO || CB
Hence
Angle CDB = DBO = 30 ---- alternate angles
Angle CBD = 180 - (120+30) = 30 -- Angle Sum Property

Your diagram will look like: