Answer
$\sqrt {25-x^2}=5~cos~θ$
Work Step by Step
$sin^2θ+cos^2θ=1$
$cos^2θ=1-sin^2θ$
$\sqrt {25-x^2}=\sqrt {25-(5~sin^2θ)^2}=\sqrt {25-25~sin^2θ}=\sqrt {25(1-sin^2θ)}=5\sqrt {cos^2θ}=±5~cos~θ$
But, since $0\lt θ\lt\frac{\pi}{2}$ (Quadrant I):
$\sqrt {25-x^2}=5~cos~θ$
