A little more detail on AKG's suggestion: You are looking for [itex]cos(\theta)[/itex] with [itex]\theta= arctan(b)[/itex] or [itex]tan(\theta)= b[/itex].
Since "[itex]tan(\theta)[/itex]" is "opposite side over near side" you can think of b as b/1. Draw a right triangle have "opposite side" b and "near side" 1. Now [itex]cos(\theta)[/itex] is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenuse and find [itex]cos(\theta)[/itex].