In the given circle with diameter AB, find the value of X.

40°

\(\angle\)ABD = \(\angle\)ACD = \({30}^\circ\)   (Angles in the same segment)

In \(\Delta \)ADB, by angle sun property we have

\(\angle\)BAD + \(\angle\)ADB + \(\angle\)ABD = \({180}^\circ\)

But we know that angle in a semi circle is \({90}^\circ\)

\(\angle\)ADB = \({90}^\circ\)

So,

x + \({90}^\circ\) + \({30}^\circ\) = \({180}^\circ\)

x = \({180}^\circ\) - \({120}^\circ\)

Therefore, x = \({60}^\circ\).