Question

In the figure given below, AOB is a diameter of the circle and $\angle AOC={110}^{\circ }$. Find $\angle BDC$.

• A. ${35}^{\circ }$
• B. ${30}^{\circ }$
• C. ${20}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A

${35}^{\circ }$

$\angle AOC+\angle COB={180}^{\circ }$ (Linear pair)

$\therefore \angle COB={180}^{\circ }-\angle AOC={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

Arc BC subtends $\angle COB$ at the centre and x at the remaining part of circle

$\therefore \angle COB=2x$

$x=\frac{1}{2}\angle COB=\frac{1}{2}\times {70}^{\circ }={35}^{\circ }$