Finding an angle inside an irregular quadrilateral

Question

I’ve been trying to solve this question for several days now and I still got no clue on how to solve it.

Find $x$

Answer 1

Let $AC\cap BD=\{E\}$.

Thus, $$\frac{\sin(103^{\circ}-x)}{\sin{x}}=\frac{ED}{EC}=\frac{ED}{AE}\cdot\frac{AE}{BE}\cdot\frac{BE}{EC}=$$ $$=\frac{\sin64^{\circ}}{\sin13^{\circ}}\cdot\frac{\sin17^{\circ}}{\sin86^{\circ}}\cdot\frac{\sin43^{\circ}}{\sin34^{\circ}}.$$ I got $x=39^{\circ}$.

I got it by the following way.

We have $$\sin103^{\circ}\cot{x}-\cos103^{\circ}=\frac{\cos26^{\circ}}{4\sin13^{\circ}\cos43^{\circ}\cos17^{\circ}}$$ or $$\cot{x}=\frac{\cos26^{\circ}}{2\sin26^{\circ}\cos43^{\circ}\cos17^{\circ}}-\tan13^{\circ}.$$ But $$\frac{\cos26^{\circ}}{2\sin26^{\circ}\cos43^{\circ}\cos17^{\circ}}-\tan13^{\circ}=$$ $$=\frac{1}{\tan26^{\circ}(\cos60^{\circ}+\cos26^{\circ})}-\tan13^{\circ}=$$ $$=\frac{1}{\frac{2\tan13^{\circ}}{1-\tan^213^{\circ}}\left(\frac{1}{2}+\frac{1-\tan^213^{\circ}}{1+\tan^213^{\circ}}\right)}-\tan13^{\circ}=$$ $$=\frac{1-\tan^413^{\circ}}{\tan13^{\circ}(3-\tan^213^{\circ})}-\tan13^{\circ}=\frac{1-3\tan^213^{\circ}}{3\tan13^{\circ}-\tan^313^{\circ}}=\cot39^{\circ},$$ which says $x=39^{\circ}.$

Answer 2

This is the construction I did with GeoGebra. The auxiliary circle is needed to construct the angle of $13°$ in $D$

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