Answer
$\sin{t} = \frac{\sqrt{13}}{7}$
$\cos{t} = -\frac{6}{7}$
$\tan{t} = -\dfrac{\sqrt{13}}{6}$
Work Step by Step
RECALL:
For the terminal point P(x, y) on a unit circle,
$\sin{t} = y, \cos{t} = x, \text{ and } \tan{t} = \frac{y}{x}, x\ne0$
Use the formulas above to obtain:
$\sin{t} = \frac{\sqrt{13}}{7}$
$\cos{t} = -\frac{6}{7}$
$\tan{t} = \dfrac{\frac{\sqrt{13}}{7}}{-\frac{6}{7}} = \dfrac{\sqrt{13}}{7} \cdot \left(-\dfrac{7}{6}\right)=-\dfrac{\sqrt{13}}{6}$
