Adventitious Quadrangle 6

Label the figure as above, where $DE = h_1$ and $AF = h_2$ are perpendicular to $BC=1$ , and $DG$ is parallel to $BC$ . Then we have:

$h_1\cot 10^\circ + h_1 \cot 20^\circ = 1$

$\begin{aligned} \implies h_1 & = \frac {\tan 10^\circ \tan 20^\circ}{\tan 10^\circ + \tan 20^\circ} = \frac {\sin 10^\circ \sin 20^\circ}{\sin 10^\circ \cos 20^\circ + \sin 20^\circ \cos 10^\circ} = \frac {\sin 10^\circ \sin 20^\circ}{\sin 30^\circ} = 2 \sin 10^\circ \sin 20^\circ \end{aligned}$

Similarly, $h_2\cot 20^\circ + h_2 \cot 60^\circ = 1$ , $\implies h_2 = \dfrac {\sin 20^\circ \sin 60^\circ}{\sin 80^\circ} = \dfrac {2\sin 10^\circ \cos 10^\circ \sin 60^\circ}{\cos 10^\circ} = \sqrt 3 \sin 10^\circ$

Then we note:

$\begin{aligned} \tan \angle ADG & = \frac {AG}{DG} \\ \tan (x - 20^\circ) & = \frac {h_2-h_1}{h_1 \cot 20^\circ - h_2\cot 60^\circ} \\ & = \frac {\sqrt 3 \sin 10^\circ - 2\sin 10^\circ \sin 20^\circ}{2\sin 10^\circ \cos 20^\circ - \sin 10^\circ} \\ & = \frac {\sqrt 3 - 2\sin 20^\circ}{2\cos 20^\circ - 1} & \small \color{#3D99F6} \text{Divide up and down by }2 \\ & = \frac {\frac {\sqrt 3} 2 - \sin 20^\circ}{\cos 20^\circ - \frac 12} \\ & = \frac {\cos 30^\circ - \cos 70^\circ}{\cos 20^\circ - \cos 60^\circ} \\ & = \frac {2 \sin 20^\circ \sin 50^\circ}{2 \sin 20^\circ \sin 40^\circ} = \frac {\sin 50^\circ}{\cos 50^\circ} = \tan 50^\circ \\ \implies x & = \boxed{70}^\circ \end{aligned}$

For the triangle above, we have:

$\angle BCD = (140 - x)^{\circ}$ and $\angle DAB = (x - 40)^{\circ}$

In addition, we have

$\dfrac{DC}{DA} \dfrac{DA}{DB} \dfrac{DB}{DC} = 1$

Now, applying the law of Sines, to triangles $\triangle DCA$ , $\triangle DAB$ , and $\triangle DBC$ , in this order, the above equation becomes,

$\dfrac{\sin 10^{\circ}}{\sin (x - 40)^{\circ}} \dfrac{\sin 20^{\circ}}{\sin 10^{\circ}} \dfrac{\sin (140-x)^{\circ}}{\sin 40^{\circ}} = 1$

which reduces to,

$\sin 20^{\circ} \sin (140-x)^{\circ} = \sin (x - 40)^{\circ} \sin 40^{\circ}$

Using the identity $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ , results in

$\sin 20^{\circ} ( \sin 140^{\circ} \cos x^{\circ} - \cos 140^{\circ} \sin x^{\circ} ) = \sin 40^{\circ} (\sin x^{\circ} \cos 40^{\circ} - \cos x^{\circ} \sin 40^{\circ} )$

Dividing through by $\cos x^{\circ}$ and re-arranging, we obtain,

$\tan x^{\circ} = \dfrac{ \sin 20^{\circ} \sin 140^{\circ} + \sin^2 40^{\circ} } { \sin 40^{\circ} \cos 40^{\circ} + \sin 20^{\circ} \cos 140^{\circ} }$

Using a scientific calculator, we can now find $\tan x^{\circ}$ , and then using the $\tan^{-1}$ function we can find $x^{\circ}$ . The answer comes to $\boxed{ 70^{\circ} }$