Angle subtended by the arc at the centre is twice the angle subtended by the same chord in the remaining part of the circle. Therefore,∠AOB = 2x Applying sine rule in Δ OAB, we get: ⇒
r
sin(90−x)
=
√3r
sin2x
⇒
r
cosx
=
√3r
2sinx⋅cosx
⇒sinx=
√3
2
⇒x=60∘ Since the sum of opposite angles in a cyclic quadrilateral is 180°, therefore,kx+x=180. 60k+60=180 60k=120⇒k=2