1. Given that $\csc x = \frac{17}{15}$ and $x$ is a positive acute angle, we need to find the value of $x$. 2. Recall that $\csc x = \frac{1}{\sin x}$, so we have: $$\sin x = \frac{1}{\csc x} = \frac{1}{\frac{17}{15}} = \frac{15}{17}$$ 3. Since $x$ is acute, $x$ lies between $0^\circ$ and $90^\circ$, so we can find $x$ by taking the inverse sine: $$x = \sin^{-1}\left(\frac{15}{17}\right)$$ 4. Calculate the value: $$x \approx \sin^{-1}(0.8824) \approx 62.0^\circ$$ Final answer: $x \approx 62.0^\circ$