Area Of A Quadrilateral!

In right $\triangle{ABD}$ we have:

$\dfrac{6}{x} = \tan(50^{\circ}) \implies x = 6\cot(50^{\circ}) \implies y = 6\csc(50^{\circ})$

Using the law of sines on $\triangle{BCD} \implies \dfrac{z}{\sin(130^{\circ})} = 6\sqrt{2}\csc(50^{\circ}) \implies$

$z = 6\sqrt{2}\csc(50^{\circ})\sin(130^{\circ})$ (will simplify all later)

and

$\dfrac{m}{\sin(5^{\circ})} = 6\sqrt{2}\csc(50^{\circ}) \implies m = 6\sqrt{2}\csc(50^{\circ})\sin(5^{\circ})$

$h = \dfrac{m}{\sqrt{2}} = 6\csc(50^{\circ})\sin(5^{\circ})$

$\implies A_{\triangle{ABD}} = \dfrac{1}{2}(6)(6\cot(50^{\circ})) = 18\cot(50^{\circ})$

and

$A_{BCD} = 18\sqrt{2}(\cot(50^{\circ})\sin(130^{\circ})\sin(5^{\circ})$

$\implies A_{ABCD} = 18(\cot(50^{\circ}) + \sqrt{2}\csc^2(50^{\circ})\sin(130^{\circ})\sin(5^{\circ}))$

and

$\cot(50^{\circ}) + \sqrt{2}\csc^2(50^{\circ})\sin(130^{\circ})\sin(5^{\circ}) =$

$\cot(50^{\circ}) + \sqrt{2}\csc^2(50^{\circ})\sin(50^{\circ})\sin(5^{\circ}) =$

$\dfrac{1}{\sin(50^{\circ})}(\cos(50^{\circ}) + \sqrt{2}\sin(5^{\circ})) =$

$\dfrac{1}{\sin(50^{\circ})}(\cos(50^{\circ}) + \sqrt{2}\sin(45^{\circ} - 40^{\circ})) =$

$\dfrac{1}{\sin(50^{\circ})}(\cos(50^{\circ}) + \cos(40^{\circ}) - \sin(40^{\circ})) =$

$\dfrac{1}{\sin(50^{\circ})}(\cos(50^{\circ}) + \sin(50^{\circ}) - \cos(50^{\circ})) = 1$

$\implies A_{ABCD} = \boxed{18}$ .