How many triangles are possible given $m\angle C={{63}^{\circ }}$ , b = 9 and c = 12?

Hint: Now we are given that b = 9, c = 12 and $\angle C={{63}^{\circ }}$ .
Now we know that according to the law of sin we have, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Complete step by step answer:
Now first consider $\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Now substituting the given values we get,
$\Rightarrow \dfrac{9}{\sin B}=\dfrac{12}{\sin {{63}^{\circ }}}$
Cross multiplying in the above equation we get,
$\Rightarrow \sin B\times 12=9\times \sin {{63}^{\circ }}$
Now substituting the value of $\sin {{63}^{\circ }}$ we get,
$\begin{align}
  & \Rightarrow \sin B\times 12=9\times 0.89. \\
 & \Rightarrow \sin B=\dfrac{9\times 0.89}{12} \\
 & \Rightarrow \sin B=0.668 \\
 & \Rightarrow B={{\sin }^{-1}}\left( 0.668 \right) \\
 & \Rightarrow B={{42}^{\circ }} \\
\end{align}$
Now we know that the sum of all angles of triangles is equal to ${{180}^{\circ }}$
Hence we get, $\angle A+\angle B+\angle C={{180}^{\circ }}$
$\begin{align}
  & \Rightarrow {{63}^{\circ }}+{{42}^{\circ }}+\angle A={{180}^{\circ }} \\
 & \Rightarrow \angle A={{75}^{\circ }} \\
\end{align}$
Hence $\angle A={{75}^{\circ }}$ .
Now consider the ratio $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$
Substituting the values in the equation we get,
$\Rightarrow \dfrac{a}{\sin {{75}^{\circ }}}=\dfrac{9}{\sin {{42}^{\circ }}}$
Now rearranging the terms we get,
$\begin{align}
  & \Rightarrow a=\dfrac{\sin {{75}^{\circ }}\times 9}{\sin {{42}^{\circ }}} \\
 & \Rightarrow a=13 \\
\end{align}$
Hence we have $a=13,b=9,c=12$ and $\angle A={{75}^{\circ }},\angle B={{42}^{\circ }}$ and $\angle C={{63}^{\circ }}$
Hence we can only form one triangle with the given values.

Note: Now note that when we have given one angle and two sides we can just draw one triangle. Similarly if we are given three sides we can just draw one triangle. Note that with just 3 angles given more than infinite triangles can be drawn.