Question

someone pls help me with this question

Answer 1

given that angle APB=150*
therefore
we know that angle APB=1/2 angle ACB
so angle ACB=150/2=75*
now
we know that
angleACB+angleDCB=180*
75*+angle DCB=180*so
angle DCB=180*-75*
angle DCB=105*
and quardiletral BCDQ is a cyclic quardiletral
and we know that the sum of its opposite side is 180*
so
angleDCB+angleDQB=180*
105*+angle DQB=180*
angle DQB=180*-105*so
angleDQB=75*
so
x=75*