Inverse Trigo

Refer to the figure below (can't upload images though, so here's the link to that image: Image for this solution).

We have by sine law:
$$ \frac{16}{\sin(120^\circ - \theta)} = \frac{8t-16}{\sin \theta} \Longrightarrow t = \frac{2 \sin \theta}{\sin(120^\circ -\theta)} + 2 $$
Differentiating with respecting to \(\theta\) gives
$$ \frac{dt}{d\theta}=2\cdot \frac{\sin(120^\circ -\theta)\cdot\cos\theta - \sin \theta\cdot -\cos(120^\circ -\theta)}{\sin^2 (120^\circ -\theta)} $$

Solving (b) is easier, since just after the turn (i.e., \(t=2\text{ hr}\)) means that \(\theta = 0^\circ\).

$$ \begin{align*} \left.\frac{dt}{d\theta}\right]_{t=2} &=2\cdot \frac{\sin(120^\circ -0^\circ)\cdot\cos0^\circ - \sin 0^\circ\cdot -\cos(120^\circ -0^\circ)}{\sin^2 (120^\circ -0^\circ)}=2.3094 \\ \left.\frac{d\theta}{dt}\right]_{t=2}&=\frac{1}{\displaystyle \left.\frac{dt}{d\theta}\right]_{t=2}} = \frac{1}{2.3094}=\color{red}{\boxed{0.4330\text{ rad}/\text{hr}}} \end{align*}$$

Now for (a) (i.e. \(t=4\text{ hr}\)), we have
$$ t=4 = \frac{2 \sin \theta}{\sin(120^\circ -\theta)} + 2 \Longrightarrow \theta = 60^\circ$$
and so
$$ \begin{align*} \left.\frac{dt}{d\theta}\right]_{t=4} &=2\cdot \frac{\sin(120^\circ -60^\circ)\cdot\cos60^\circ - \sin 60^\circ\cdot -\cos(120^\circ -60^\circ)}{\sin^2 (120^\circ -60^\circ)}=2.3094 \\ \left.\frac{d\theta}{dt}\right]_{t=4}&=\frac{1}{\displaystyle \left.\frac{dt}{d\theta}\right]_{t=4}} = \frac{1}{2.3094}=\color{red}{\boxed{0.4330\text{ rad}/\text{hr}}} \end{align*}$$