## Precalculus (6th Edition) Blitzer

$257$
Here, we have $\dfrac{x}{ \tan 40^{\circ}} = \dfrac{x}{ \tan 20^{\circ}}-400$ or, $x= \dfrac{400}{\dfrac{1}{\tan 20^{\circ}}-\dfrac{1}{\tan 40}}$ or, $x= \dfrac{400}{2.7475-1.1918}$ Simplify the above equation: so, $x=257$