In triangle VWX shown, VX = sqrt13 and VW = 3. What is tan V?

Hey there!

First use pythagoras to gain the length of \(WX\)

\(WX=\sqrt{(\sqrt{13})^2-3^3}=2\)

Now we can take the tangent of \(V\)

\(tan(V)=\frac{\sqrt{(\sqrt{13})^2-3^3}}{3}=\frac{2}{3}\)

Hope this helped :)

Edit: Had \(13^2\) instead of \(\sqrt{13}^2\) oops