Answer
(a) Reference Number: $\pi/6$
(b) Terminal point: $(\sqrt 3/2, 1/2)$
Work Step by Step
1. $13\pi/6 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$.
$$\frac{13\pi}{6} - 2\pi = \frac{13\pi}{6} -\frac{12\pi}{6} = \frac{\pi}{6}$$
2. $\pi/6$ is in Quadrant I, thus, the Reference Number is $\pi/6$:
$$t^- = \frac{\pi}{6}$$
3. According to the Table 1, the terminal point for $\pi/6$ is $(\sqrt 3/2, 1/2)$. Since the terminal point must be in Quadrant I, the x-coordinate is positive, and the y-coordinate is also positive. Therefore, the terminal point for $t = 13\pi/6$ is $(\sqrt 3/2, 1/2)$