It's All Angles !

Let $|\overline {AB}|=|\overline {CD}|=x, |\overline {BD}|=y$ . Then

$\dfrac{x}{\sin \theta}=\dfrac{y}{\sin 80\degree}\implies \dfrac{x}{y}=\dfrac{\sin \theta}{\sin 80\degree}, \dfrac{x}{\sin (\theta-20\degree)}=\dfrac{y}{\sin 20\degree}\implies \dfrac{x}{y}=\dfrac{\sin (\theta-20\degree)}{\sin 20\degree}$ . This implies $\sin \theta\sin 20\degree=\sin (\theta-20\degree)\sin 80\degree\implies \theta=\tan^{-1} \left (\dfrac{\sin 20\degree\sin 80\degree}{\cos 20\degree\sin 80\degree-\sin 20\degree}\right) =\boxed {30\degree}$ .

Using the law of cosines on $\triangle{ABC}$ with included angle $C$ we obtain:

$x^2 = 2l^2(1 - \cos(20^{\circ})) = 4l^2\sin^2(10^{\circ}) \implies x = 2l\sin(10^{\circ})$

Using the law of cosines on $\triangle{BCD}$ with included angle $C$ we obtain:

$b^2 = l^2(1 + 4\sin^2(10^{\circ}) - 4\sin(10^{\circ})\cos(20^{\circ}))$

Using law of sines on $\triangle{ABD}$ we obtain:

$\dfrac{2l\sin(10^{\circ})}{\sin(\theta)} = \dfrac{b}{\sin(80^{\circ})} \implies$ $b = \dfrac{2l\sin(10^{\circ})\sin(80^{\circ})}{\sin(\theta)} \implies$

$b^2 = \dfrac{4l^2\sin^2(10^{\circ})\sin^2(80^{\circ})}{\sin^2(\theta)} \implies$

$\dfrac{4\sin^2(10^{\circ})\sin^2(80^{\circ})}{\sin^2(\theta)} = 1 + 4\sin^2(10^{\circ}) - 4\sin(10^{\circ})\cos(20^{\circ}) \implies$

$\sin^2(\theta) = \dfrac{4\sin^2(10^{\circ})\sin^2(80^{\circ})}{1 + 4\sin^2(10^{\circ}) - 4\sin(10^{\circ})\cos(20^{\circ})} \implies$

$\sin(\theta) = \dfrac{2\sin(10^{\circ})\sin(80^{\circ})}{\sqrt{1 + 4\sin^2(10^{\circ}) - 4\sin(10^{\circ})\cos(20^{\circ})}} =$

$\dfrac{\sin(20^{\circ})}{\sqrt{1 + 4\sin^2(10^{\circ}) - 4\sin(10^{\circ})\cos(20^{\circ})}} = 0.5 \implies \theta = \boxed{30^{\circ}}$