Rate of change of angle of elevation

Set up a right triangle with the fixed horizontal distance $x = 60$ ft from the person to the point below the kite, height $h$, and angle of elevation $\theta$. Since $x$ is constant, use $\tan\theta = \dfrac{h}{60}$. Differentiating both sides with respect to $t$ gives $\sec^2\theta \, \dfrac{d\theta}{dt} = \dfrac{1}{60}\dfrac{dh}{dt}$.

At the given instant the string is $100$ ft, $x=60$ ft, so by the Pythagorean theorem $h = \sqrt{100^2-60^2} = 80$ ft. Then $\sec\theta = \dfrac{100}{60}$, so $\sec^2\theta = \dfrac{10000}{3600} = \dfrac{25}{9}$. Substituting $dh/dt = 5$ ft/s: $\dfrac{25}{9}\dfrac{d\theta}{dt} = \dfrac{5}{60} = \dfrac{1}{12}$, so $\dfrac{d\theta}{dt} = \dfrac{9}{12\cdot 25} = \dfrac{3}{100}$ rad/s.