In right triangle $MNO$, $\tan{M}=\frac{5}{4}$, $OM=8$, and $\angle O = 90^\circ$.  Find $MN$.  Express your answer in simplest radical form.
I know that the tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side, so if $\tan{M}=\frac{5}{4}$, then $\frac{NO}{OM}=\frac{5}{4}$. I can cross-multiply to get $4NO=5OM$. Since $OM=8$, I can substitute that and get $4NO=40$. Dividing by 4, I get $NO=10$. Now I have the lengths of the two legs of the right triangle, and I want to find the hypotenuse $MN$. I can use the Pythagorean theorem, which says that $MN^2=OM^2+NO^2$. Plugging in the values, I get $MN^2=8^2+10^2$. Simplifying, I get $MN^2=64+100$. Adding, I get $MN^2=164$. Taking the square root of both sides, I get $MN=\sqrt{164}$. I can simplify the radical by looking for perfect square factors. I notice that 164 is divisible by 4, which is a perfect square. I can write 164 as $4\times 41$. Then I can use the property that $\sqrt{ab}=\sqrt{a}\sqrt{b}$. I get $\sqrt{164}=\sqrt{4}\sqrt{41}$. Simplifying, I get $\sqrt{164}=2\sqrt{41}$. 