Answer
$\sin{x} = \dfrac{12}{13}$
$\tan{x}= -\dfrac{12}{5}$
Work Step by Step
Since $x$ lies in the second quadrant, $\sin{x}$ is positive.
Using the formula:
$$\sin{x}=\sqrt{1-\cos^2{x}}$$
$$\sin{x}=\sqrt{1-\left(-\dfrac{5}{13} \right)^2}$$
$$\sin{x} = \dfrac{12}{13}$$
$\because \tan{x}=\dfrac{\sin{x}}{\cos{x}}$
$\therefore \tan{x}=\dfrac{\dfrac{12}{13}}{-\dfrac{5}{13}}$
$\tan{x}= -\dfrac{12}{5}$
