Solve the triangle.
$$\mathrm{A}={51}^{\circ },\mathrm{C}={67}^{\circ },\mathrm{C}=11$$
$$B={◻}^{\circ }$$
$a=◻$ (Do not round until the final answer. Then round to one decimal place as needed.)
$b=◻$ (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: $B=\boxed{{\displaystyle {62}^{\circ }}}$

Step 13 ：Final Answer: $a=\boxed{{\displaystyle 9.3}}$

Step 14 ：Final Answer: $b=\boxed{{\displaystyle 10.6}}$