Solve the triangle.
\[
\mathrm{A}=51^{\circ}, \mathrm{C}=67^{\circ}, \mathrm{C}=11
\]
\[
B=\square^{\circ}
\]
$a=\square$ (Do not round until the final answer. Then round to one decimal place as needed.)
$b=\square$ (Do not round until the final answer. Then round to one decimal place as needed.)

Step 1 ：To solve the triangle, we first find the missing angle \( B \) using the fact that the sum of angles in a triangle is \( 180^{\circ} \).

Step 2 ：Since \( A = 51^{\circ} \) and \( C = 67^{\circ} \), we calculate \( B \) as follows: \( B = 180^{\circ} - A - C \).

Step 3 ：Substitute the given values: \( B = 180^{\circ} - 51^{\circ} - 67^{\circ} \).

Step 4 ：Calculate the value of \( B \): \( B = 62^{\circ} \).

Step 5 ：Now, we use the Law of Sines to find the missing sides \( a \) and \( b \). The Law of Sines states that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).

Step 6 ：To find \( a \), we rearrange the formula: \( a = c \cdot \frac{\sin A}{\sin C} \).

Step 7 ：Substitute the given values: \( a = 11 \cdot \frac{\sin 51^{\circ}}{\sin 67^{\circ}} \).

Step 8 ：Calculate the value of \( a \) and round to one decimal place: \( a = 9.3 \).

Step 9 ：To find \( b \), we rearrange the formula: \( b = c \cdot \frac{\sin B}{\sin C} \).

Step 10 ：Substitute the given values: \( b = 11 \cdot \frac{\sin 62^{\circ}}{\sin 67^{\circ}} \).

Step 11 ：Calculate the value of \( b \) and round to one decimal place: \( b = 10.6 \).

Step 12 ：Final Answer: \( B = \boxed{62^{\circ}} \)

Step 13 ：Final Answer: \( a = \boxed{9.3} \)

Step 14 ：Final Answer: \( b = \boxed{10.6} \)