Question

In the figure at left, O is the centre of the circle, $\angle AOC={100}^o$ and side AB is produced to D.
Find (i) $\angle CBD$ (ii) $\angle ABC$

Answer 1

$\angle AOC=100°$

$ABCE$ is a cyclic quadrilateral.

$\angle AEC=\frac{1}{2}\angle AOC=50°$ [ Since as angle made by the chord at the surface of the circle is half of the angle made by the same chord at the center]

In quadrilateral $ABCE$;

$\angle AEC+\angle ABC=180°⟹\angle ABC=130°$ [ Since in a cyclic quadrilateral sum of diagonal angle is 180°]

$\angle ABC+\angle CBD=180°$

$⟹\angle CBD=50°$

Hence;

$\angle CBD=50°$ and

$\angle ABC=130°$