In triangle $PQR$, we have $\angle P = 90^\circ$, $QR = 15$, and $\tan R = 5\cos Q$.  What is $PQ$?
I notice that this is a right triangle, so I can use the Pythagorean theorem and the trigonometric ratios. I also notice that the given equation relates the tangent of one acute angle to the cosine of the other, which suggests using the complementary angle identity $\cos Q = \sin (90^\circ - Q) = \sin R$. Substituting this into the equation, I get $\tan R = 5\sin R$, which I can rewrite as $\sin R / \cos R = 5\sin R$. Dividing both sides by $\sin R$, I get $1 / \cos R = 5$, or $\cos R = 1/5$. Now I can use the Pythagorean theorem to find $PR$, since $QR^2 = PQ^2 + PR^2$. Plugging in the values, I get $15^2 = PQ^2 + PR^2$, or $225 = PQ^2 + PR^2$. I can also use the cosine ratio to find $PR$, since $\cos R = PR / QR$. Plugging in the values, I get $1/5 = PR / 15$, or $PR = 3$. Substituting this into the Pythagorean equation, I get $225 = PQ^2 + 3^2$, or $225 = PQ^2 + 9$. Subtracting 9 from both sides, I get $216 = PQ^2$. Taking the square root of both sides, I get $PQ = \sqrt{216}$. I can simplify this radical by factoring out the largest perfect square, which is 36. So $PQ = \sqrt{36 \cdot 6} = \sqrt{36} \cdot \sqrt{6} = 6\sqrt{6}$. 