<s>[INST] Triangular sides 
In a triangle the least angle is $45º$ and the tangents of the angle are in arithmetic progression. If its area is $27\text{cm}^2$, find the length of the sides.

I tried to solve the problem in this way
Let the smallest angle in the triangle $ABC$ (say) be $A=45º$, where $\tan A$, $\tan B $ and $\tan C$ are in $AP$, then according to question
$$\begin{array}
2\tan B=\tan A+\tan C\\
\implies 2\tan B=\frac{\sin A\cos C+\cos A\sin C}{\cos A\cos C}\\
\implies 2\tan B=\frac{\sin B}{\cos A\cos C}\\
\implies \cos B=2\cos A\cos C
\end{array}$$
As we know that $\cos A=\frac{1}{\sqrt2}$ hence
$$\cos B=\sqrt2\cos C$$
squaring both sides would give 
$$\cos^2B=2\cos^2C$$
Changing the cosines to sines we get
$$1-\sin^2B=2-2\sin^2C$$
$$=2\sin^2C-\sin^2B=1$$
Now as we already know that $\sin C=\frac{2\Delta}{ab}$ and similarly $\sin B=\frac{2\Delta}{ac}$ where $\Delta$ is the area of the triangle and $a,b,c$ the sides of the respective triangle.
But now when I put the sine formula in the equation and take $\Delta=27\text{cm}^2$ I get stuck with the product of $abc$ what should I do to get the sides.
 [/INST] HINT:
$\tan A=1,$
If $\tan B=p,\tan C=\tan(135^\circ -B)=\dfrac{p+1}{p-1}$
We have $$\tan A+\tan C=2\tan B\implies1+\dfrac{p+1}{p-1}=2p$$
Can you take it from here using 
$$27=\dfrac12bc\sin A\iff bc=54\sqrt2$$
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