Angle subtended by a chord AB on the centre of the circle is 140 degree. What will be the angle subtended by the chord AB on the major are of the circle?

70 degree

The angle subtended by an arc at the center of a circle is double the size of the angle subtended by the same arc at the circle circumference

Take any point Q on the circumference and join AQ and QC.

Therefore, \(\angle\)AOC = 2 x \(\angle\)AQC

= \(\angle\)AQC = \(\frac{1}{2}\) x \(\angle\)AOC = \(\frac{1}{2}\) x \({140}^\circ\) = \({70}^\circ\)

Now, \(\angle\)CBP = \(\angle\)AQC [Exterior angle of cyclic quadrilateral]

= \(\angle\)CBP = \({70}^\circ\)