Bridge to Nowhere

A large number of angles can be easily obtained and some are shown in the image above. To find $x$ , we can arbitrarily set $\overline{DE}=1$ . From there we can go over the bridge to $\overline{CE}$ using the law of sines three times for the triangles $\triangle AED, \triangle ABE, \triangle BCE$ .

$\frac{\overline{AE}}{\sin(41)}=\frac{1}{\sin(54)}, \frac{\overline{BE}}{\sin(42)}=\frac{\overline{AE}}{\sin(43)}, \frac{\overline{CE}}{\sin(51)}=\frac{\overline{BE}}{\sin(44)}$

$\overline{CE}=\frac{\sin(51)}{\sin(44)}\times\frac{\sin(42)}{\sin(43)}\times\frac{\sin(41)}{\sin(54)}=0.89011$

In the $\triangle CDE$ the $\angle CED=95^\circ$ and the sides next to it are 1 and 0.89011. $\overline{CD}$ can be obtained from the law of cosines as

$\overline{CD}=\sqrt{1+0.89011^2-2\times0.89011\times \cos(95)}=1.3955$

The law of sines applied to the $\triangle CDE$ provides $x$ as

$x= \arcsin(\frac{\sin(95)}{1.3955})=45.55^\circ$