## Precalculus (6th Edition) Blitzer

The exact value is $\frac{5}{12}$.
Let us assume $\theta$ represents the angle. $\theta ={{\sin }^{-1}}\left( \frac{5}{13} \right)$ Then, $\sin \theta =\frac{5}{13}$ Since, $\sin \theta$ is positive, $\theta$ is in the first quadrant. Now, using the Pythagorean identity, we get \begin{align} & {{x}^{2}}+{{y}^{2}}={{r}^{2}} \\ & {{x}^{2}}+{{5}^{2}}={{13}^{2}} \\ & {{x}^{2}}=169-25 \\ & x=12 \end{align} Thus, \begin{align} & \tan \left( {{\sin }^{-1}}\left( \frac{5}{13} \right) \right)=\tan \theta \\ & =\frac{x}{y} \\ & =\frac{5}{12} \end{align}