Solve the triangle. \[ A=35^{\circ}, \quad B=35^{\circ}, \quad c=2 \] \( C=\square^{\circ} \)

In a triangle, the sum of angles A, B, and C is always \(180^{\circ}\). Given that \(A = 35^{\circ}\) and \(B = 35^{\circ}\), we can calculate angle \(C\): \[ C = 180^{\circ} - A - B = 180^{\circ} - 35^{\circ} - 35^{\circ} = 110^{\circ}. \] Now, using the Law of Sines, we can find the other sides, \(a\) and \(b\). The Law of Sines states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. \] Since \(c = 2\), we can find \(a\) as follows: \[ \frac{a}{\sin 35^{\circ}} = \frac{2}{\sin 110^{\circ}}. \] We know that \(\sin 110^{\circ} = \sin 70^{\circ} \approx 0.9397\) and \(\sin 35^{\circ} \approx 0.5736\). So, \[ a = \frac{2 \cdot \sin 35^{\circ}}{\sin 110^{\circ}} \approx \frac{2 \cdot 0.5736}{0.9397} \approx 1.22. \] To find \(b\): \[ \frac{b}{\sin 35^{\circ}} = \frac{2}{\sin 110^{\circ}} \implies b \approx 1.22. \] In summary, the triangle has: \( C = 110^{\circ} \), \( a \approx 1.22 \), and \( b \approx 1.22 \).