1. **State the problem:** We need to find side $g$ (EF) in triangle EGF where angle $G = 101^\circ$, side $EG = 5$, and side $GF = 4$. 2. **Formula used:** The Law of Cosines states: $$g^2 = EG^2 + GF^2 - 2 \times EG \times GF \times \cos(G)$$ This formula relates the lengths of sides of a triangle to the cosine of one of its angles. 3. **Substitute known values:** $$g^2 = 5^2 + 4^2 - 2 \times 5 \times 4 \times \cos(101^\circ)$$ 4. **Calculate each term:** $$g^2 = 25 + 16 - 40 \times \cos(101^\circ)$$ 5. **Evaluate $\cos(101^\circ)$:** Since $101^\circ$ is obtuse, $\cos(101^\circ)$ is negative. Using a calculator: $$\cos(101^\circ) \approx -0.1908$$ 6. **Plug in the cosine value:** $$g^2 = 25 + 16 - 40 \times (-0.1908)$$ $$g^2 = 41 + 7.632$$ $$g^2 = 48.632$$ 7. **Find $g$ by taking the square root:** $$g = \sqrt{48.632} \approx 6.97$$ 8. **Round to the nearest tenth:** $$g \approx 7.0$$ **Final answer:** $$\boxed{7.0}$$