In the figure given below, P andQ are centres of two circles, intersecting at B and C , and ACD is a straight line.

If `/_ APB = 150^(@)` and `/_ BQD = x^(@)` , find the value of x.

Explanation

Join BC.
We know that the angle made by than arc at the centre is twice the angle made by the arc at the point on the remaining part of the circle.
`:. /_ ACB = (1)/(2) /_ APB = ((1)/(2) xx 150^(@))= 75^(@)`
Now, ACD being a straight line, we have
`/_ ACB + /_ BCD = 180^(@)`
`implies 75^(@) + /_ BCD = 180^(@) implies /_ BCD = 180^(@) - 75^(@) = 105^(@)`
Similarly, for the second circle, we have
`/_ BCD = (1)/(2) ` reflex `/_ BQD`.
[ angle made by the major arc BFD at the centre`=2 /_ BCD`]
`implies 105^(@) = (1)/(2 ) (360^(@) -x^(@)) = x= 150`
Hence, `x= 150`