right triangle

Let $|\overline {AC}|=1$ . Then $|\overline {AB}|=\cos 50\degree, |\overline {BC}|=\sin 50\degree$ . $|\overline {BE}|=|\overline {BD}|\tan (70\degree-x)$ . Now, $\dfrac{\cos 50\degree-|\overline {BE}|}{\sin 30\degree}=\dfrac{1}{\sin 100\degree}\implies \cos 50\degree-|\overline {BE}|=\dfrac{1}{2\sin 100\degree}, \dfrac{\sin 50\degree-|\overline {BD}|}{\sin 30\degree}=\dfrac{1}{\sin 110\degree}\implies \sin 50\degree-|\overline {BD}|=\dfrac{1}{2\sin 110\degree}$ . So, $\cos 50\degree-\dfrac{1}{2\sin 100\degree}=\left (\sin 50\degree-\dfrac{1}{2\sin 110\degree}\right )(\tan (70\degree-x)\implies \tan (70\degree-x)=\tan 30\degree\implies x=\boxed {40\degree}$ .