1. **Problem Statement:** Find the value of $\sin K$ in the right triangle with vertices $K$, $J$, and $I$, where side $KJ = \sqrt{14}$, side $KI = 7$, and $J$ is the right angle. 2. **Recall the definition of sine in a right triangle:** $$\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}$$ 3. **Identify the sides relative to angle $K$:** - Opposite side to angle $K$ is $JI$ (unknown length). - Adjacent side to angle $K$ is $KJ = \sqrt{14}$. - Hypotenuse is $KI = 7$. 4. **Find the length of side $JI$ using the Pythagorean theorem:** $$KI^2 = KJ^2 + JI^2$$ $$7^2 = (\sqrt{14})^2 + JI^2$$ $$49 = 14 + JI^2$$ $$JI^2 = 49 - 14 = 35$$ $$JI = \sqrt{35}$$ 5. **Calculate $\sin K$:** $$\sin K = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JI}{KI} = \frac{\sqrt{35}}{7}$$ 6. **Simplify the fraction:** $$\sin K = \frac{\sqrt{35}}{7} = \frac{\cancel{\sqrt{7}} \times \sqrt{5}}{\cancel{7}} = \frac{\sqrt{5}}{\sqrt{7}}$$ 7. **Rationalize the denominator:** $$\sin K = \frac{\sqrt{5}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{35}}{7}$$ 8. **Calculate the decimal value:** $$\sin K \approx \frac{5.916}{7} \approx 0.845$$ 9. **Round to the nearest hundredth:** $$\sin K \approx 0.85$$ **Final answer:** $$\boxed{0.85}$$