enter image description here

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?


  • $\begingroup$ 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

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$


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 $CDB$)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.