IF L is the circumcentre of $$\triangle$$XYZ and angle X is $$40^\circ$$, then the value of $$\angle$$YZL is:

Given, L is the circumcentre of $$\triangle$$XYZ and $$\angle$$YXZ = $$40^\circ$$

Angle subtended by an arc of the circle at centre is twice the angle subtended by the arc at any point on the circle.

$$\Rightarrow$$ Angle subtended by the arc YZ at L is twice the angle subtended by the arc YZ at X.

$$\Rightarrow$$ $$\angle$$YLZ = 2$$\angle$$YXZ

$$\Rightarrow$$ $$\angle$$YLZ = $$80^\circ$$

In $$\triangle$$LYZ, YL = ZL

Angles opposite to equal sides are equal.

$$\Rightarrow$$ $$\angle$$YZL = $$\angle$$ZYL

Also, $$\angle$$YZL + $$\angle$$ZYL + $$\angle$$YLZ = $$180^\circ$$

$$\Rightarrow$$ $$\angle$$YZL + $$\angle$$YZL + $$80^\circ$$ = $$180^\circ$$

$$\Rightarrow$$ 2$$\angle$$YZL = $$100^\circ$$

$$\Rightarrow$$ $$\angle$$YZL = $$50^\circ$$

Hence, the correct answer is Option A