Question

In the diagram, O is the centre of the circle. The angles CBD is equal to:

Answer 1

In $△AOC$,

$\angle AOC={50}^o$

And $AO=OC$,

Therefore,$\angle OAC=\angle OCA$

$\angle OAC=\angle OCA=\frac{{180}^o-{50}^o}{2}={65}^o$

Therefore $\angle ABC=\frac{{50}^o}{2}={25}^o$ (alternate segment theorem )

Now $\angle CBD=180-25={155}^o$ (linear pair)

So $\angle CBD={155}^o$