Answer
a.
$\overline{t}=0$
b.
$\displaystyle \overline{t}=\frac{\pi}{4}$
c.
$\displaystyle \overline{t}=\frac{\pi}{6}$
d.
$\overline{t}=4-\pi\approx 0.86$
Work Step by Step
The reference number associated with the real number $t$ is the shortest distance along the unit circle between the terminal point determined by $t$ and the x-axis.
For each t, find each terminal point on the unit circle (positive=counterclockwise) and associate it with the terminal point of some t between 0 and $ 2\pi$
If the terminal point "lands" in quadrants II, III or IV,
choose the symmetric terminal number ($\pm\pi$) in quadrant I$:$
t in Q.II $\Rightarrow \overline{t}=\pi-t$
t in Q.III$\Rightarrow \overline{t}=t-\pi$
t in Q.IV$\Rightarrow \overline{t}=2\pi-t$
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a. $ t=9\pi$ has the same terminal point as $\pi$,
its reference number is $\pi-\pi=0$
b. $ \displaystyle \frac{-5\pi}{4}$ (clockwise from 0), has the same terminal point as$ \displaystyle \frac{3\pi}{4}$ ( in Q.II )
its reference number is $\displaystyle \pi-\frac{3\pi}{4}=\frac{\pi}{4} $
c. The terminal point of $t=\displaystyle \frac{25\pi}{6}=4\pi+\frac{\pi}{6}$ is the same terminal point as $\displaystyle \frac{\pi}{6}$ (Q.I),
its reference number is $\displaystyle \frac{\pi}{6}$
d. The terminal point of 4 is in Q.III, ($\pi\approx$3.14),
its reference number is $4-\pi\approx 4-3.14=0.86$