$
\eqalign{
& \frac{1}
{{\sqrt 3 }} = \frac{{50 - x}}
{x} \cr
& x = \left( {50 - x} \right) \cdot \sqrt 3 \cr
& x = 50\sqrt 3 - x\sqrt 3 \cr
& x + x\sqrt 3 = 50\sqrt 3 \cr
& x\left( {1 + \sqrt 3 } \right) = 50\sqrt 3 \cr
& x = \frac{{50\sqrt 3 }}
{{1 + \sqrt 3 }} = 75 - 25\sqrt 3 \cr}
$

De laatste stap behoeft wellicht nog enige toelichting:

$
\eqalign{
& \frac{{50\sqrt 3 }}
{{1 + \sqrt 3 }} = \cr
& \frac{{50\sqrt 3 }}
{{1 + \sqrt 3 }} \cdot \frac{{1 - \sqrt 3 }}
{{1 - \sqrt 3 }} = \cr
& \frac{{50\sqrt 3 - 150}}
{{1 - 3}} = \cr
& \frac{{50\sqrt 3 - 150}}
{{ - 2}} =; \cr
& 75 - 25\sqrt 3 \cr}
$

O ja... de worteltruuk.