Question

In the given figure, 0 is the centre of the circle. If $\angle BAD={75}^o$ and BC = CD, find $\angle BOD$,$\angle BCD$,$\angle OBD$

Answer 1

An angle subtended by any arc on centre of circle is twice to that subtended at any other point on circle,

$\Rightarrow$By arc BD. $\angle BOD=2\times \angle BAD={150}^o$

and $\angle BCD=\frac{1}{2}\left(\angle BOD\right)={75}^o$, In $\Delta BOD,OB=OD$

$\Rightarrow \angle OBD=\angle ODB$ and $\angle OBD+\angle ODB+\angle BOD={180}^o$

$\Rightarrow 2\angle OBD=180-150$

$\Rightarrow \angle OBD={15}^o$.