1. **State the problem:** We are given a triangle ABC with angles $\angle A = 98.4^\circ$, $\angle B = 24.6^\circ$, and side $AC = 400$ units. We need to find side $x = AB$, which is opposite angle $B$. 2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. So, $$\angle C = 180^\circ - 98.4^\circ - 24.6^\circ = 57^\circ.$$ 3. **Apply the Law of Sines:** The Law of Sines states $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C},$$ where side $a$ is opposite angle $A$, side $b$ opposite angle $B$, and side $c$ opposite angle $C$. Here, side $AC = 400$ is opposite angle $B = 24.6^\circ$, and side $x = AB$ is opposite angle $C = 57^\circ$. So, $$\frac{x}{\sin 57^\circ} = \frac{400}{\sin 24.6^\circ}.$$ 4. **Solve for $x$:** $$x = \frac{400 \times \sin 57^\circ}{\sin 24.6^\circ}.$$ Calculate the sines: $$\sin 57^\circ \approx 0.8387, \quad \sin 24.6^\circ \approx 0.4161.$$ So, $$x = \frac{400 \times 0.8387}{0.4161} \approx \frac{335.48}{0.4161} \approx 806.5.$$ 5. **Final answer:** The length of side $x$ is approximately **806.5** units.