Solve the triangle. \[ \mathrm{A}=120^{\circ}, \mathrm{C}=35^{\circ}, \mathrm{c}=120 \] \( \mathrm{~B}=25^{\circ} \) \( \mathrm{a} \approx \square \)

To solve for side \( a \) in triangle ABC, we can use the Law of Sines, which states that \(\frac{a}{\sin A} = \frac{c}{\sin C}\). Given that \( A = 120^\circ \), \( C = 35^\circ \), and \( c = 120 \), we first find \( a \): \[ \frac{a}{\sin(120^\circ)} = \frac{120}{\sin(35^\circ)} \] Calculating \(\sin(120^\circ)\) and \(\sin(35^\circ)\): - \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \frac{\sqrt{3}}{2}\). - \(\sin(35^\circ) \approx 0.5736\). Plug these values into the equation: \[ \frac{a}{\frac{\sqrt{3}}{2}} = \frac{120}{0.5736} \] Now, multiply both sides by \(\frac{\sqrt{3}}{2}\): \[ a = \frac{120 \cdot \frac{\sqrt{3}}{2}}{0.5736} \] Calculating further, we find: \[ a \approx \frac{60\sqrt{3}}{0.5736} \approx 166.4 \] So, \( a \approx 166.4 \).