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I'm trying to find the value of $x$. For that, I need to get some help. Let's call $\angle A = 2\beta $, $\angle B = \alpha$ and $\angle C = 60$ We know that

$$\triangle ABC = 180 $$

$$\angle A +\angle B +\angle C = 180$$

Hence we have

$$\beta + \alpha = 60$$

My apologies if I'm wrong. Can you explain how to proceed?

Regards

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  • $\begingroup$ Is it safe to assume that all the central angles are right? $\endgroup$ – Rhys Hughes May 20 '18 at 12:59
  • $\begingroup$ @RhysHughes Yes, feel free! :) $\endgroup$ – Busi May 20 '18 at 12:59
  • $\begingroup$ Then $x$ is $30^\circ$. $\endgroup$ – poyea May 20 '18 at 13:01
  • $\begingroup$ How did you get from $\angle A+\angle B+\angle C=180$ to $\beta+\alpha=60$? I get $2\beta+\alpha=180-60=120$. $\endgroup$ – Barry Cipra May 20 '18 at 13:35
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Assuming that the shape is a kite and that all central angles are right angles; the solution is pretty straightforward.

If $\triangle ABC$ $= 180^\circ$

Then $\triangle DCB=180^\circ$

Focus on the right triangle $DC$. Since we're assuming that the central angles are right angles, $ x = 180-90-60$

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If the angles in the centre are right angles and therefore $90^0$, then angle $CDM$, (where $M$ is the centre) is $180-90-60=30^0$, and since $CDM=x; x=30^0$ too. (The line $AD$ bisects the angle $CDB$)

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  • $\begingroup$ What if the central angles aren't right? Is there a way to proceed? $\endgroup$ – Busi May 20 '18 at 13:10
  • $\begingroup$ Take $ABCD$ as a regular four-sided shape. $ABC=DBC=ACB=DCB=60^0$, then $360^0=240^0+4x\to x=30^0$ $\endgroup$ – Rhys Hughes May 20 '18 at 13:12

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