Question

In the given figure, $AOB$ is a diameter of circle and $\angle AOC={104}^o$. Find $\angle BDC$.

Answer 1

Join $AD$

Here $angleADC=\frac{1}{2}\angle AOC=\frac{1}{2}\times {104}^{\circ }={52}^{\circ }$

(Angle at the centre is double the angle at the circumference subtended by the same chord)

Also, $\angle ADB={90}^{\circ }$

(Angle in a semi-circle is a right angle)

$\therefore \angle BDC={90}^{\circ }-\angle ADC={90}^{\circ }-{55}^{\circ }={35}^{\circ }$