Answer
$\sqrt {9-x^2}=3~sin~θ$
Work Step by Step
$sin^2θ+cos^2θ=1$
$cos^2θ=1−sin^2θ$
$\sqrt {9-x^2}=\sqrt {9-(3~cosθ)^2}=\sqrt {9-9~cos^2θ}=\sqrt {9(1-cos^2θ)}=3\sqrt {sin^2θ}=±3~sin~θ$
But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I):
$\sqrt {9-x^2}=3~sin~θ$
