AB is the diameter of a circle with centre O. C and D are two points on the circumference of the circle on either side of AB, such that ∠CAB = 42^o and ∠ABD= 57^o. What is difference (in degrees) between the measures of ∠CAD and ∠CBD ?

\({30}^\circ\)

\(\angle\)ACB = \({90}^\circ\)  [as AB is diameter]

So, in \(\Delta \)ACB

\(\angle\)CBA = \({180}^\circ\) - \(\angle\)ACB - \(\angle\)CAB

⇒ \(\angle\)CBA = \({180}^\circ\) - \({90}^\circ\) - \({42}^\circ\)

⇒ \(\angle\)CBA = \({48}^\circ\)

So,

\(\angle\)CBD = \(\angle\)CBA - \(\angle\)ABD

⇒ \(\angle\)CBD = \({48}^\circ\) - \({57}^\circ\)

⇒ \(\angle\)CBD = \({105}^\circ\)

Again,

ADBC is a cyclic quadrilateral

So, \(\angle\)CAD = \({180}^\circ\) - \(\angle\)CBD

⇒ \(\angle\)CAD = \({180}^\circ\) - \({105}^\circ\)

⇒ \(\angle\)CAD = \({75}^\circ\)

Difference \({105}^\circ\) - \({75}^\circ\)

⇒ \({30}^\circ\)

Therefore, the difference (in degrees) between the measures of ∠CAD and ∠CBD is \({30}^\circ\)