A and B are $100$ m apart. A runner R runs at a speed of 6 m/s from A to B. Given that $BC = 50$ m and $\angle CBA = \frac{\pi}{3}$. Let $BR$ = $x$ m and $\angle BCR = \theta$. Find the rate of change of $\theta$ when $R$ is $60$ m away from $A$.
Sine law shows:
$$x = \frac{50\sin\theta}{\sin(\frac{\pi}{3} + \theta)} $$
then \begin{align} \frac{d\theta}{dt} &= \frac{dx}{dt}\frac{d\theta}{dx} \\ &= -6 \cdot \frac{\sin^2(\frac{\pi}{3} + \theta)}{25\sqrt{3}} \end{align} Also, $$40 = \frac{50\sin\theta_0}{\sin(\frac{\pi}{3} + \theta_0)}$$ but I couldn't find the exact value of $\sin^2(\frac{\pi}{3} + \theta_0)$ from this. How should I proceed? Thanks in advance.
You have $$\dfrac{\sin (60+\alpha)}{\sin \alpha}=\dfrac{5}{4}$$
$$\dfrac{\sin 60 \cos \alpha+\cos 60 \sin \alpha}{\sin \alpha}=\dfrac{5}{4}$$
$$\Rightarrow \cot \alpha = \dfrac{\sqrt{3}}{2}$$
$$\Rightarrow \sin \alpha = \dfrac{1}{\sqrt{1+\cot ^2 \alpha}}=\dfrac{2}{\sqrt 7}$$
$$\therefore \sin (60+\alpha) = \dfrac{5}{4} \cdot \dfrac{2}{\sqrt 7}$$