In \( \triangle A B C \), side \( \overline{B C} \) is extended through \( C \) to \( D \). If \( \mathrm{m} \angle A=30^{\circ} \) and \( \mathrm{m} \angle A C D=110^{\circ} \), what is the longest side of \( \triangle A B C \) ? \( \begin{array}{ll}\text { (1) } \overline{A C} & \text { (3) } \overline{A B} \\ \text { (2) } \overline{B C} & \text { (4) } \overline{C D}\end{array} \)

To determine the longest side of triangle \( \triangle ABC \), we first need to find the measures of the angles in the triangle. Given: - \( \mathrm{m} \angle A = 30^\circ \) - \( \mathrm{m} \angle ACD = 110^\circ \) Since \( D \) is on the extension of side \( BC \), we can find \( \angle ACB \) using the fact that angles on a straight line sum to \( 180^\circ \): \[ \angle ACD + \angle ACB = 180^\circ \] Substituting the known value: \[ 110^\circ + \angle ACB = 180^\circ \] Solving for \( \angle ACB \): \[ \angle ACB = 180^\circ - 110^\circ = 70^\circ \] Now, we can find \( \angle ABC \) using the fact that the sum of the angles in a triangle is \( 180^\circ \): \[ \angle A + \angle ABC + \angle ACB = 180^\circ \] Substituting the known values: \[ 30^\circ + \angle ABC + 70^\circ = 180^\circ \] Simplifying this: \[ \angle ABC + 100^\circ = 180^\circ \] Thus, \[ \angle ABC = 180^\circ - 100^\circ = 80^\circ \] Now we have all the angles in triangle \( ABC \): - \( \angle A = 30^\circ \) - \( \angle ABC = 80^\circ \) - \( \angle ACB = 70^\circ \) To find the longest side of triangle \( ABC \), we use the property that the longest side is opposite the largest angle. The angles are: - \( \angle A = 30^\circ \) (opposite side \( BC \)) - \( \angle ABC = 80^\circ \) (opposite side \( AC \)) - \( \angle ACB = 70^\circ \) (opposite side \( AB \)) The largest angle is \( \angle ABC = 80^\circ \), which is opposite side \( AC \). Therefore, the longest side of triangle \( ABC \) is: \[ \overline{AC} \] Thus, the answer is \( \boxed{1} \).