In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110°, find ∠BDC.

Join AD.

We know that,

Angle at the center is double the angle at the circumference subtended by the same chord.

∠ADC = $\dfrac{1}{2}$∠AOC = $\dfrac{1}{2}$ x 110° = 55°.

Also, we know that

Angle in the semi-circle is a right angle.

∠ADB = 90°

From figure,

∠BDC = ∠BDA - ∠ADC = 90° - 55° = 35°.

Hence, ∠BDC = 35°.