### Instruction: 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.### Response: 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 $\sqrt{164}=2\sqrt{41}$.