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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.

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2 Answers 2

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Hint:

$\triangle AOC$ is isosceles with \begin{align} \angle OCA = \angle AOC &= \tfrac12\,(180^\circ-20^\circ) =80^\circ . \end{align}

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  • $\begingroup$ I tried to prove $\triangle AOC$ is isosceles but failed.So please help me to prove this. $\endgroup$ Apr 21, 2020 at 8:01
  • $\begingroup$ @Supriyo Banerjee: can you see that $\triangle ACD$ is isosceles? $\endgroup$
    – g.kov
    Apr 21, 2020 at 8:21
  • $\begingroup$ yeah, $\angle CDA$ & $\angle ACD$ are both $70^o$. So $\triangle ACD$ is isosceles. $\endgroup$ Apr 21, 2020 at 8:25
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    $\begingroup$ @Supriyo Banerjee: well, then $|AD|=|AC|=|AO|$. $\endgroup$
    – g.kov
    Apr 21, 2020 at 8:27
  • $\begingroup$ Yeah,then AC =AD & also AD=AO hence AO=AC .Thank you so much. $\endgroup$ Apr 21, 2020 at 8:27
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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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  • $\begingroup$ FD & AB straight line can't intersect each other but you consider P as intersection of AB and FD.Also you wrote FD is drawn with angle $30^o$ with FB at F but how do you know that angle DFB = $30^o$ $\endgroup$ Apr 21, 2020 at 7:21
  • $\begingroup$ Sorry it is a typo. I meant FG. I edited my answer. $\endgroup$
    – sirous
    Apr 21, 2020 at 8:25
  • $\begingroup$ what about P point? $\endgroup$ Apr 21, 2020 at 9:07
  • $\begingroup$ Of course, P is the intersection of lines FG and AB. $\endgroup$
    – sirous
    Apr 21, 2020 at 15:55

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