is there a way to solve it with non-right angled trig? Because a way I would use required just algebra and drawing your own right angled tringles (from the left end of the 18m down, and the same with the right end).
To calculate the width of the river (let's call it w), consider when it's both closed and open.
Closed: w = 2x (because x + x)
Open: using the right angled triangles you drew up and rearranging it, w = 2xcos50+ 18.
Then just sub in w into 2x, then solve the width from there