Answer
See below.
Work Step by Step
By Pythagorean's Theorem we know that $a^2+b^2=10^2=100$ and that $a^2+(b+9)^2=17^2=289\\a^2+b^2+18b+81=289\\100+18b=208\\18b=108\\b=6$
Thus $a^2+6^2=100\\a^2=64\\a=\pm8$
But $a$ must be positive, so $a=8,b=6$
