In $∆ABC,\angle B=\frac{\pi }{4},\angle C=\frac{\pi }{3}.$ If the area of the triangle is $54+18\sqrt{3}$ sq. units, then $a=$

Given, $\angle B=\frac{\mathrm{\pi }}{4},\angle C=\frac{\mathrm{\pi }}{3}$

In a $△ABC$, $\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle A=\frac{5\mathrm{\pi }}{12}={75}^{\circ }$

$\sin A=\sin {75}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}},\sin B=\sin {45}^{\circ }=\frac{1}{\sqrt{2}} \& \sin C=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$

Using Sine Rule,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

$\Rightarrow \sin A=ak, \sin B=bk \& \sin C=ck$

$\Rightarrow \frac{\sin A}{k}=a,\frac{ \sin B}{k}=b \& \frac{\sin C}{k}=c$

Replace constant $\frac{1}{k}$ with $t$

$\Rightarrow t\sin A=a, t\sin B=b \& t\sin C=c$

$\Rightarrow a=t \sin A=\frac{\sqrt{3}+1}{2\sqrt{2}}t \dots \dots \dots \left(1\right)$

$\Rightarrow b=t \sin B=\frac{1}{\sqrt{2}}t$

$\Rightarrow c=t \sin C=\frac{\sqrt{3}}{2}t$

Now, Area of triangle $=\frac{1}{2}bc\times \sin A=\frac{1}{2}\times \frac{1}{\sqrt{2}}t\times \frac{\sqrt{3}}{2}t\times \frac{\sqrt{3}+1}{2\sqrt{2}}$

$\Rightarrow 54+18\sqrt{3}=\frac{1}{2}\times \frac{3+\sqrt{3}}{8}{t}^2$

$\Rightarrow 18\left(3+\sqrt{3}\right)=\frac{1}{2}\times \frac{3+\sqrt{3}}{8}{t}^2$

$\Rightarrow t^2=18\times 2\times 8$

$\Rightarrow t=12\sqrt{2}$

Substituting value of $t$ in equation $\left(1\right)$

$\therefore a=\frac{\sqrt{3}+1}{2\sqrt{2}}\times 12\sqrt{2}=6\left(\sqrt{3}+1\right)$