1. **State the problem:** We have a triangle with sides 70, 52, and 48 units. We want to find the angle $x$ opposite the side of length 70. 2. **Formula used:** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $c$ is the side opposite angle $C$. 3. **Apply the formula:** Here, $c=70$, $a=52$, $b=48$, and $C=x$. $$70^2 = 52^2 + 48^2 - 2 \times 52 \times 48 \times \cos(x)$$ 4. **Calculate squares:** $$4900 = 2704 + 2304 - 4992 \cos(x)$$ 5. **Simplify the right side:** $$4900 = 5008 - 4992 \cos(x)$$ 6. **Isolate the cosine term:** $$4900 - 5008 = -4992 \cos(x)$$ $$-108 = -4992 \cos(x)$$ 7. **Divide both sides by -4992:** $$\cancel{-108} \div \cancel{-4992} = \cos(x)$$ $$\cos(x) = \frac{108}{4992}$$ 8. **Simplify the fraction:** $$\cos(x) = \frac{9}{416} \approx 0.02163$$ 9. **Find the angle $x$ using inverse cosine:** $$x = \cos^{-1}(0.02163) \approx 88.8^\circ$$ **Final answer:** $x \approx 88.8^\circ$ (nearest tenth) This corresponds to choice C.