I posted the previous reply before I realized that you'd posted this twice. Sorry.
View attachment 8139Unknown: a and b.
Known: all the angles and d.
My work so far:
Really? The image below looks to be the work of whoever sent this to you, who addressed you as "Jon".
This image is also small, but I'm pretty sure the other helper said the following:
We have \(\displaystyle a\, =\, \dfrac{P\, \cos(\theta)}{3}\) with \(\displaystyle O\, =\, 2\, (a\, +\, d)\, \cos(\theta).\)
The partial derivation \(\displaystyle \dfrac{\partial}{\partial\, a}\, \left(2\, (a\, +\, d)\, \cos(\theta)\right)\, =\, 2\, \cos(\theta)\)
Hence \(\displaystyle a\) is equivalent to \(\displaystyle a\, =\, \dfrac{O}{2\, \cos(\theta)}\, -\, d\)
So we have the following equality:
. . . . .\(\displaystyle \dfrac{1}{3}\, \left(O\, \cos(\theta)\right)\, =\, \dfrac{O}{2\, \cos(\theta)}\, -\, d\)
Assuming that \(\displaystyle d,\, O,\) and \(\displaystyle \theta\) are positive
. . . . .\(\displaystyle O\, \left(\cos(\theta)\right)\, =\, -6d\, \cos(\theta)\)
Thus
. . . . .\(\displaystyle O\, =\, -\dfrac{6d\, \cos(\theta)}{\cos(2\theta)\, -\, 2}\)
Have you taken calculus and differential equations?