Answer
$\sqrt {9x^2+25}=5~sec~θ$
Work Step by Step
$sec^2θ=1+tan^2θ$
$\sqrt {9x^2+25}=\sqrt {(3x)^2+25}=\sqrt {(5~tan~θ)^2+25}=\sqrt {25~tan^2θ+25}=\sqrt {25(tan^2θ+1)}=5\sqrt {sec^2θ}=±5~sec~θ$
But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I):
$\sqrt {9x^2+25}=5~sec~θ$
