We have to find the angle x
I have tried using these theorems, but failed.
1. The angle at the centre is twice the angle at the circumference.
2. The exterior angle of a triangle is equal to the sum of the two opposite interior angle.
$\endgroup$2 Answers
$\begingroup$Hint:
$\triangle AOC$ is isosceles with\begin{align} \angle OCA = \angle AOC &= \tfrac12\,(180^\circ-20^\circ) =80^\circ . \end{align}
$\endgroup$ 5 $\begingroup$Consider the drawing. Extend AO to make diameter AD. Extend AC to make chord AE. Clearly AE and AB are symmetric about AD and are equal. FG is drawn with angle $30^o$ with FB at F, so its intersect with AB make a point which is symmetric with C about AD. $\angle CAO=\angle DAB=20^o$. So points C, O and P(intersection of FD and AB) are on one circle(are cyclic) so $AO=AC$, that is triangle AOC is isosceles and $X=\angle AC0=80^o$.
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