Finding angles in a parallelogram without trigonometry  
I'm wondering whether it's possible to solve for $x^{\circ}$ in terms of $a^{\circ}$ and $b^{\circ}$ given that $ABCD$ is a parallelogram. In particular, I'm wondering if it's possible to solve it using only "elementary geometry". I'm not sure what "elementary geometry" would usually imply, but I'm trying to solve this problem without trigonometry.
Is it possible? Or if it's not, is there a way to show that it isn't solvable with only "elementary geometry" techniques?

The example by alex.jordan does finish the matter, and similar ones may be constructed. We have an angle
$$ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $$
and we wish to know whether $ x = \frac{\theta}{\pi}  $ is the root of an equation with rational coefficients. 
Well,
$$ e^{i \theta} = \sqrt{\frac{12}{13}} + i   \sqrt{\frac{1}{13}}  $$ 
Next, $\cos 2 \theta = 2 \cos^2 \theta - 1 = \frac{11}{13}.$ So, by Corollary 3.12 on page 41 of NIVEN we know that $2 \theta$ is not a rational multiple of $\pi.$ So, neither is $\theta,$ and
 $$ x = \frac{\theta}{\pi}  $$
is irrational.
Now, the logarithm is multivalued in the complex plane. We may choose 
$$ \log(-1) = \pi i.  $$ With real $x,$ we have chosen
$$ (-1)^x = \exp(x \log(-1)) =  \exp(x\pi i) = \cos \pi x + i \sin \pi x.  $$
With our $ x = \frac{\theta}{\pi},  $ we have 
$$   (-1)^x =  e^{i \pi x}  =  e^{i \theta} = \sqrt{\frac{12}{13}} + i   \sqrt{\frac{1}{13}}     $$
The right hand side is algebraic.
The Gelfond-Schneider Theorem, Niven page 134, says that if $\alpha,\beta$ are nonzero algebraic numbers, with $\alpha \neq 1$ and $\beta$ not a real rational number, then any value of $\alpha^\beta$ is transcendental.
Taking $\alpha = -1$ and $\beta = x,$ which is real but irrational. We are ASSUMING that $x$ is algebraic over $\mathbb Q.$ The assumption, together with Gelfond-Schneider, says that $ (-1)^x$ is transcendental. However, we already know that $   (-1)^x = \sqrt{\frac{12}{13}} + i   \sqrt{\frac{1}{13}}     $ is algebraic. This contradicts the assumption. So $x = \theta / \pi$ is transcendental, with $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $
