Answer
(a) Reference Number: $\pi/3$
(b) Terminal Point: $(- 1/2, -\sqrt 3/2)$
Work Step by Step
1. $16\pi/3 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$.
$$\frac{16\pi}{3} - 2\pi = \frac{16\pi}{3} - \frac{6\pi}{3} = \frac{10\pi}{3}$$ $$\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}$$
2. $4\pi/3$ is in Quadrant III, thus, the Reference Number is given by the equation:
$$t^- = \frac{4\pi}{3} - \pi = \frac{4\pi}{3} - \frac{3\pi}{3} = \frac{\pi} 3$$
3. According to the Table 1, the terminal point for $\pi/3$ is $(1/2, \sqrt 3/2)$. In Quadrant III: x-coordinate is negative and y-coordinate is also negative. Therefore, the terminal point is: $(- 1/2, -\sqrt 3/2)$
