## Precalculus (6th Edition)

$a = \pm \sqrt{c^2-b^2}$
Subtract $b^2$ to both side of the equation to obtain: $$a^2+b^2-b^2=c^2-b^2 \\a^2=c^2-b^2$$ Take the square root of both sides of the equation to obtain: $$\sqrt{a^2}=\pm \sqrt{c^2-b^2} \\a = \pm \sqrt{c^2-b^2}$$