$h = 10(t - 2) \sin 60^\circ$
$h = 10(t - 2)\left( \dfrac{\sqrt{3}}{2} \right)$
$h = 5\sqrt{3}(t - 2)$
$x = 10(t - 2) \cos 60^\circ$
$x = 10(t - 2)(1/2)$
$x = 5(t - 2)$
$b = 20 + x = 20 + 5(t - 2)$
$b = 20 + 5t - 10 = 5t + 10$
$b = 5(t + 2)$
$\theta = \arctan \dfrac{h}{b}$
$\theta = \arctan \dfrac{5\sqrt{3}(t - 2)}{5(t + 2)}$
$\theta = \arctan \dfrac{\sqrt{3}(t - 2)}{t + 2}$
$\dfrac{d\theta}{dt} = \dfrac{\dfrac{(t + 2)\sqrt{3} - \sqrt{3}(t - 2)(1)}{(t + 2)^2}}{1 + \left[ \dfrac{\sqrt{3}(t - 2)}{t + 2} \right]^2}$
$\dfrac{d\theta}{dt} = \dfrac{\dfrac{t\sqrt{3} + 2\sqrt{3} - t\sqrt{3} + 2\sqrt{3}}{(t + 2)^2}}{1 + \dfrac{3(t - 2)^2}{(t + 2)^2}}$
$\dfrac{d\theta}{dt} = \dfrac{\dfrac{4\sqrt{3}}{(t + 2)^2}}{\dfrac{(t + 2)^2 + 3(t - 2)^2}{(t + 2)^2}}$
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(t + 2)^2 + 3(t - 2)^2}$
(a) 4 hours after the start, t = 4
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(4 + 2)^2 + 3(4 - 2)^2}$
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{36 + 12}$
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{48}$
$\dfrac{d\theta}{dt} = \frac{1}{12}\sqrt{3} \, \text{ rad/sec}$ answer
(b) Just after the turn, t = 2
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(2 + 2)^2 + 3(2 - 2)^2}$
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{16 + 0}$
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{16}$
$\dfrac{d\theta}{dt} = \frac{1}{4}\sqrt{3} \, \text{ rad/sec}$ answer