A geometry problem by A Former Brilliant Member

All angles in the following solution are in degree measure.

$4\sin(50) - \sqrt{3}\tan(50) = \dfrac{4\sin(50)\cos(50) - \sqrt{3}\sin(50)}{\cos(50)} =$

$\dfrac{2\sin(100) - \sqrt{3}\sin(50)}{\cos(50)} = \dfrac{2\sin(80) - \sqrt{3}\sin(50)}{\cos(50)}$ ,

where the identities $2\sin(x)\cos(x) = \sin(2x)$ and $\sin(x) = \sin(180 - x)$ were used. Now

$\sin(80) = \sin(30 + 50) = \sin(30)\cos(50) + \cos(30)\sin(50) = \dfrac{1}{2}\cos(50) + \dfrac{\sqrt{3}}{2}\sin(50)$ ,

and so $\dfrac{2\sin(80) - \sqrt{3}\cos(50)}{\cos(50)} = \dfrac{(\cos(50) + \sqrt{3}\sin(50)) - \sqrt{3}\sin(50)}{\cos(50)} = \boxed{1}$ .