1. **State the problem:** We need to find the value of $\cos W$ in a right triangle with vertices $X$, $W$, and $Y$, where the right angle is at $X$. The side $XY$ (opposite to angle $W$) has length $\sqrt{14}$, and the hypotenuse $YW$ has length $\sqrt{78}$. 2. **Recall the cosine definition:** In a right triangle, $\cos$ of an angle is the ratio of the length of the adjacent side to the hypotenuse. For angle $W$, the adjacent side is $XW$, and the hypotenuse is $YW$. 3. **Find the length of side $XW$ using the Pythagorean theorem:** $$YW^2 = XW^2 + XY^2$$ Substitute the known values: $$\sqrt{78}^2 = XW^2 + \sqrt{14}^2$$ $$78 = XW^2 + 14$$ 4. **Solve for $XW^2$:** $$XW^2 = 78 - 14 = 64$$ 5. **Find $XW$:** $$XW = \sqrt{64} = 8$$ 6. **Calculate $\cos W$:** $$\cos W = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XW}{YW} = \frac{8}{\sqrt{78}}$$ 7. **Rationalize the denominator:** $$\cos W = \frac{8}{\sqrt{78}} \times \frac{\sqrt{78}}{\sqrt{78}} = \frac{8\sqrt{78}}{78}$$ 8. **Simplify the fraction:** $$\frac{8\sqrt{78}}{78} = \frac{\cancel{2} \times 4 \sqrt{78}}{\cancel{2} \times 39} = \frac{4\sqrt{78}}{39}$$ 9. **Approximate the value:** Calculate $\sqrt{78} \approx 8.83$, so $$\cos W \approx \frac{4 \times 8.83}{39} = \frac{35.32}{39} \approx 0.91$$ **Final answer:** $$\boxed{\cos W \approx 0.91}$$