In the given figure, *P* and *Q* are centres of two circles intersecting at *B* and *C*. *ACD* is a straight line. Then, ∠*BQD* =

#### Solution

Consider the circle with the centre ‘*P*’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

`angleACB = (angleAPB )/2`

`=(150°)/2`

`angleACB` = 75°

Since ‘*ACD*’ is a straight line, we have

`angleACB + angleBCD` = 180°

`angleBCD = 180° - angleACB`

= 180° - 75°

`angleBCD ` = 105°

Now let us consider the circle with centre ‘*Q*’. Here let ‘*E*’ be any point on the circumference along the major arc ‘*BD*’. Now ‘*CBED*’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

`angleBCD + angleBED` = 180°

`angleBED = 180° - angleBCD`

= 180° - 105°

`angleBED` = 75°

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, now we have

`angleBQD = 2 angleBED`

=2(75°)

`angleBQD` = 150°

Hence, the measure of `angleBQD` is 150° .