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?
 A: Let me rephrase your question first: Can the angle $x$ in the diagram be expressed in terms of angles $a$ and $b$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers, and no trig or inverse trig functions?  (For example, is $x$ an expression like $(a+2b)^{3/2}$?)
Here is strong evidence that the answer is no.  If $a=\frac{\pi}{2}$ and $b=\frac{\pi}{3}$ (the two "simplest" angles to play with) then allowing ourselves access to trigonometry, we can show that $$x=2\arctan\left(\sqrt{13}-2\sqrt{3}\right)=\arctan\left(\frac{\sqrt{3}}{6}\right)$$
So you would have to believe that this expression is equal to a combination of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers.  I highly doubt this.  (The transcendence of $\pi$ and $\arctan$ might be used to even prove that it is not possible - but that's just my conjecture.)
A: 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) $
A: In the case $a = \pi/2$, $x = \arccos \left( 2\,{\frac {\sin \left( b \right) }{\sqrt {4-3\, 
\cos^2 \left( b \right)   }}} \right)$.  This is not an algebraic function of $b$, because its derivative is $\frac{dx}{db} = \frac{2}{3 \cos^2 b - 4}$ for $-\pi/2 < b < \pi/2$, and $\cos(b)$ is not an algebraic function. 
A: On taking into account the triangles ACP and ABP, where P is the point where the diagonals intercept, we have, from sines' law:  $\sin{x}/\sin{a} = \sin{\theta}/\sin{b}$, $\theta$ being the angle between the smaller diagonal and the line AB. 
Then $\sin{\theta} = \sin{x} \, \dfrac{\sin{b}}{\sin{a}}$. 
Now, note that, in triangle ACP, the sum of the inner angles is $\,x+a+b+\theta = 180^{\circ}$, hence $\sin{\theta} = \sin{(x+a+b)}$. 
Therefore: $\sin{x} \, \dfrac{\sin{b}}{\sin{a}} = \sin{(x+a+b)} = \sin{x} \, \cos{(a+b)} + \sin{(a+b)} \, \cos{x}$. 
This implies that $\cot{x} = \dfrac{\sin{b}}{\sin{a} \, \sin{(a+b)}} - \cot{(a+b)}$.
