Answer
a) The tangential speed of the person in Ecuador is $465.74m/s$
b) $\theta=70.53^o$
Work Step by Step
a) The Earth finishes 1 revolution, or $2\pi$ radians, in $23.9h=8.604\times10^4s$. Its angular speed is $$\omega=\frac{2\pi}{8.604\times10^4}=7.3\times10^{-5}rad/s$$
The Earth's radius $r_E=6.38\times10^6m$. The tangential speed of the person in Ecuador, therefore, is $$v_E=\omega r_E=465.74m/s$$
b) Take the tangential speed of a person in place A to be $v_A$, where $v_A=1/3v_E$. We have $$\frac{v_A}{v_E}=\frac{\omega r_A}{\omega r_E}=\frac{r_A}{r_E}=\frac{1}{3}$$
As the figure below shows, we can find the angle $\theta$ by using its cosine: $$\theta=\cos^{-1}\Big(\frac{r_A}{r_E}\Big)=70.53^o$$
