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Please see diagram below.

I am trying to find an expression for $θ$ that makes the area of the bottom region $(2a)$ equal to the sum of the other two areas $(a)$. Assume this is a unit circle, so $r=1$.

Diagram not drawn to scale Diagram not drawn to scale

Diagram #2

The area of circular segment $a$ of a unit circle is: $A_a = \frac{θ_1 - sin(θ_1)}{2}$
Note that this $θ_1$ is not the $θ$ shown in the first diagram, but that: $θ_1 = π - θ$

The area of circular segment $2a$ of a unit circle is $A_{2a} = \frac{θ_2 - sin(θ_2)}{2}$

Let: $A_a = \frac{A_{2a}}{2}$
Then we have: $\frac{θ_1 - sin(θ_1)}{2} = \frac{θ_2 - sin(θ_2)}{4}$

We know that: $θ_1 + \frac{θ_2}{2} = π$
Solving for $θ_1$ we get: $θ_1 = π - \frac{θ_2}{2}$

After substituting we get:
$\frac{(π - \frac{θ_2}{2}) - sin(π - \frac{θ_2}{2})}{2} = \frac{θ_2 - sin(θ_2)}{4}$

If I could solve this for $θ_2$, finding $θ$ would be trivial.
The Inscribed Angle Theorem says that: $θ = \frac{θ_2}{2}$

I was able to get what looks like a numerical approximation using wolframalpha, but I can't find a closed form solution. I think $θ$ works out to be: $1.2473826912977909520$ radians or $71.46976364903345985$ degrees

Brovidio
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    Hi, welcome to MSE! What have you tried? Give your thoughts on the problem, or context about where you found it. – ndhanson3 Jul 21 '21 at 15:57
  • Assume radius r and base a (they are related with $\theta$) - from there calculate the sectors - you get two equations with two unknowns. – Moti Jul 21 '21 at 18:38
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    A numerical solution is often the best you can get for this kind of equations where the same unknown appears both inside and outside a trig function. Having closed-form solutions for that is the exception rather than the rule. – Troposphere Jul 22 '21 at 14:54
  • It's easy to see that the answer is between $60^\circ$ (where the sum of the two "side" sectors is bigger than the "bottom" sector) and $90^\circ$ (where the sum of the two "side" sectors is smaller than the "bottom" sector) so at least you have a sanity check on your answer. – NickD Jul 22 '21 at 16:40
  • @Henry, your equation is not the same as the OP's equation. – NickD Jul 22 '21 at 16:42
  • @NickD in his defense his comment was posted before I added my equations. – Brovidio Jul 22 '21 at 16:43
  • Oh, I doubt he needs any defending :-) My comment was directed more to latecomers (like me) who might be puzzled by the discrepancy. – NickD Jul 22 '21 at 16:48
  • @NickD, Oh, I just looked at his profile. I'm new here. I wasn't aware of his celebrity :) more likely that MY equation is wrong lol. – Brovidio Jul 22 '21 at 16:54
  • No, I'm pretty sure you got it right. – NickD Jul 22 '21 at 17:10
  • If you are talking about me, I did not know I was a celebrity. In any case I can be wrong, and looking again at this particular case I probably am. So I will delete my earlier comment. The answer in the question of $1.24738269$ looks reasonable. It is also a solution of $2\theta +\sin(\theta)(1-\cos(\theta))=\pi$ – Henry Jul 22 '21 at 17:31

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