Question

In the figure, O is the centre of the circle. Find $\angle CBD$.

Answer 1

Arc AC subtends $\angle AOC$ at the centre and $\angle APC$ at the remaining part of the circle.

$\therefore \angle APC=\frac{1}{2}\angle AOC$$=\frac{1}{2}\times {100}^{\circ }={50}^{\circ }$

$\because$ APCB is a cyclic quadrilateral,

$\therefore \angle APC+\angle ABC={180}^{\circ }$

$\Rightarrow {50}^{\circ }+\angle ABC={180}^{\circ }$

$\Rightarrow \angle ABC={180}^{\circ }-{50}^{\circ }={130}^{\circ }$

But $\angle ABC+\angle CBD={180}^{\circ }\left(LinearPair\right)$

$\Rightarrow {130}^{\circ }+\angle CBD={180}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

$\therefore \angle CBD={50}^{\circ }$