since those two triangles are similar by AA
then use proportions to get "x"
thus [tex]\bf \cfrac{x+4}{16}=\cfrac{x-08}{WU}[/tex] hmmm wait just a second? what the dickens is WU anyway?
well, let's take a look at the triangle on the right-side
is a right-triangle, has a 90° angle
so, just use the pythagorean theorem to get WU
[tex]\bf c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\qquad
\begin{cases}
c=WY\\
a=UY\\
b=WU
\end{cases}\\\\
-----------------------------\\\\
\sqrt{(WY)^2-(UY)^2}=WU[/tex]
which will give you [tex]\bf \cfrac{x+4}{16}=\cfrac{x-08}{\boxed{\sqrt{(WY)^2-(UY)^2}}}[/tex]
solve for "x"
