Use the Law of Sines to solve for the remaining side(s) and angle(s) of the triangle if $\alpha=14^{\circ}, \beta=19^{\circ}, a=9$. Round to two decimal places. As in the text, $(\alpha, a),(\beta, b)$ and $(\gamma, c)$ are angle-side opposite pairs. If no such triangle exists, enter DNE in each answer box. \[ \begin{array}{l} \gamma=\square \text { degrees } \\ b=\square \\ c=\square \end{array} \]

Step 1 ：Given the angles \( \alpha = 14^{\circ} \) and \( \beta = 19^{\circ} \), and the side \( a = 9 \).

Step 2 ：First, convert the angles from degrees to radians because the trigonometric functions work with radians.

Step 3 ：Calculate \( \gamma \) using the formula \( \gamma = \pi - \alpha - \beta \) because the sum of angles in a triangle is \( \pi \) radians.

Step 4 ：Calculate \( b \) and \( c \) using the Law of Sines: \( b = a \cdot \frac{\sin(\beta)}{\sin(\alpha)} \) and \( c = a \cdot \frac{\sin(\gamma)}{\sin(\alpha)} \).

Step 5 ：Convert \( \gamma \) back to degrees.

Step 6 ：Round \( \gamma \), \( b \), and \( c \) to two decimal places.

Step 7 ：Final Answer: \( \gamma = \boxed{147.0} \) degrees, \( b = \boxed{12.11} \), and \( c = \boxed{20.26} \).