Answer
$\sqrt {49-x^2}=7~cos~θ$
Work Step by Step
$sin^2θ+cos^2θ=1$
$cos^2θ=1−sin^2θ$
$\sqrt {49-x^2}=\sqrt {49-(7~sin~θ)^2}=\sqrt {49-49~sin^2θ}=\sqrt {49(1-sin^2θ)}=7\sqrt {cos^2θ}=±7~cos~θ$
But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I):
$\sqrt {49-x^2}=7~cos~θ$
