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,
\[\angle BDC\text{ }=\text{ }{{90}^{o}}-\angle ADC\text{ }=\text{ }{{90}^{o}}-\text{ }{{55}^{o}}\]
\[\angle BDC\text{ }=\text{ }{{35}^{o}}\]