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To solve the problem, I followed the following steps:

Is it possible to find a closed form for $x$?


where, $$x:= \angle OBC,\beta:=\angle ABC ,\alpha:=\angle OAC, \gamma:=\angle OCA ,\theta:=\angle BAC=\angle ACB$$

Mathematica says that,

$x\approx 0.033921 \approx 1.94353^\circ\,$

Is there another method to find $x$?

  • $\begingroup$ I would like to thank @Batominovski for helping me take the steps above. $\endgroup$ – Elementary Dec 3 '18 at 16:11
  • $\begingroup$ You can use use here compound angle formula and eleminate as far as possible the variables. $\endgroup$ – priyanka kumari Dec 3 '18 at 16:14

If everything except $x$ is known, $$ \frac{\sin(x)}{\sin(\beta-x)}=c$$ reduces to $$ \tan(x) = \frac{c \sin(\beta)}{1+ c \cos(\beta)}$$

  • 1
    $\begingroup$ Teacher, Are the steps correct I have taken to solve the problem? $\endgroup$ – Elementary Dec 3 '18 at 16:20
  • 1
    $\begingroup$ What steps have you taken? $\endgroup$ – Robert Israel Dec 3 '18 at 16:59
  • $\begingroup$ >$$\frac{\sin(x)}{\sin(\beta-x)}=\frac{\sin(\alpha)\,\sin(\theta-\gamma)}{\sin(\gamma)\,\sin(\theta-\alpha)}$$ $\endgroup$ – Elementary Dec 3 '18 at 18:20

Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:

  1. Find. $\angle BAC$, $\angle BCA$ using the fact that the triangle is isosceles
  2. Find $\angle BAO$ and $\angle BCO$
  3. Find $\angle AOB$ and $\angle COB$
  4. Relate $\angle BAO$, $\angle AOB$, $\angle BCO$ and $\angle AOC$ with each other using the sine rule.

You can get a solution of the form $\frac{\sin(x+c_1)}{\sin(x+c_2)}=\frac{\sin(c_3)}{\sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.

  • $\begingroup$ Can you give me a link? $\endgroup$ – Elementary Dec 3 '18 at 19:24
  • $\begingroup$ I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :) $\endgroup$ – Jam Dec 3 '18 at 19:31
  • $\begingroup$ what is $c_1,...c_4$ $\endgroup$ – Elementary Dec 3 '18 at 19:37

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