In the following figure, P and Q are centres of two circles. The circles are intersecting at points A and B. PA produced on both the sides meets the circles at C and D. If ∠CPB=100°, then find the value of x.

100

In the given figure,

\(\angle\)CPB = \({100}^\circ\)

So, \(\angle\)APB = \({80}^\circ\) [As, \(\angle\)APB = Supplementary Angle of \(\angle\)CPB ]

Now, in \(\Delta \)APB,

AP = PB = radius of the smaller circle

So, \(\angle\)PAB = \(\angle\)PBA

\(\angle\)PAB = [(\({180}^\circ\) - \({80}^\circ\))/2]

⇒ \(\angle\)PAB = \({50}^\circ\)

Now,

\(\angle\)DAB = \({180}^\circ\) - \({50}^\circ\)

⇒ \(\angle\)DAB = \({130}^\circ\)  [As, \(\angle\)DAB =  Supplementary Angle of \(\angle\)PAB ]

As ABRD is a cyclic quadrilateral

So, \(\angle\)BRD = \({180}^\circ\) - \({130}^\circ\) = \({50}^\circ\)

Now,

According to the concept,

\(\angle\)BQD = 50 x 2 = \({100}^\circ\) i.e; x

Therefore, value of x is \({100}^\circ\).