Solution:

Join \[AD\]

So, we get

\[\angle ADC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{110}^{o}}~=\text{ }{{55}^{o}}\]

[Angle at the centre is double the angle at the circumference subtended by the same chord]

And, we know that

\[\angle ADB\text{ }=\text{ }{{90}^{o}}\]

[Angle in the semi-circle is a right angle]

Hence,

\[\angle BDC\text{ }=\text{ }{{90}^{o}}-\angle ADC\text{ }=\text{ }{{90}^{o}}-\text{ }{{55}^{o}}\]

\[\angle BDC\text{ }=\text{ }{{35}^{o}}\]