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