## Trigonometry 7th Edition

a. $-\dfrac{4}{3}$
$\sin{\theta} = \dfrac{y}{r} = \dfrac{4}{5}$ $\therefore y = 4 \hspace{20pt} r =5$ $x^2 = r^2-y^2 \\ x = \pm \sqrt{r^2-y^2} \\ = \pm \sqrt{5^2-4^2}$ $x = \pm 3$ $\because \theta$ terminates in $QII$ $\therefore x = -3$ $\tan{\theta} = \dfrac{y}{x} = \boxed{-\dfrac{4}{3}}$