Answer
$\dfrac{12}{5}$
Work Step by Step
Suppose $\theta =\cos^{-1} (\dfrac{5}{13})$
This gives: $\cos \theta=\dfrac{5}{13}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ y=\sqrt {r^2-x^2}=\sqrt {(13)^2-(5)^2}=12$
Therefore, we have $\tan \theta =\tan [\cos^{-1} (\dfrac{5}{13})]=\dfrac{12}{5}$
