trigonometry - In an triangle the least angle is $45^circ$ and the tangents of the angles are in $A.P.$If its area be $27$ sq.cm.Find the lengths of its sides.

In an triangle the least angle is $45^\circ$ and the tangents of the angles are in $A.P.$If its area be $27$ sq.cm.Find the lengths of its sides.

---

Let $A$ be the smallest angle.$\angle A=45^\circ$.$\tan A,\tan B,\tan C$ are in arithmatic progression.

$2\tan B=\tan A+\tan C$

$2\frac{\sin B}{\cos B}=\frac{\sin B}{\cos A\cos C}$

$\cos A\cos C=\frac{\cos B}{2}$

$\frac{1}{\sqrt2}\cos C=\frac{\cos B}{2}$

$\frac{\cos B}{\cos C}=\sqrt2$

I am stuck here.Please help.

Answer

almagest has provided a good answer. This answer uses your idea.

Using that $B+C=180^\circ-45^\circ=135^\circ$, you can have
$$\frac{\cos B}{\cos C}=\sqrt 2\quad\Rightarrow\quad \cos B=\sqrt 2\ \cos(135^\circ-B)=-\cos B+\sin B,$$
i.e.$$2\cos B=\sin B$$

Squaring the both sides gives
$$4(1-\sin^2B)=\sin^2B\quad\Rightarrow\quad \sin B=\frac{2}{\sqrt 5},\quad \cos B=\frac{1}{\sqrt 5}\tag1$$
because $B\le 90^\circ$, and so we have
$$\sin C=\sin(135^\circ-B)=\frac{3}{\sqrt{10}}\tag2$$

By the way,
$$27=\frac 12bc\sin A\quad\Rightarrow\quad bc=54\sqrt 2$$
and so using the law of sines with $(1)(2)$ gives
$$2R\sin B\cdot 2R\sin C=54\sqrt 2\quad\Rightarrow\quad R=\frac{3\sqrt 5}{\sqrt 2}$$where $R$ is the circumradius.

Thus,
$$a=2R\sin A=3\sqrt 5,\quad b=2R\sin B=6\sqrt 2,\quad c=2R\sin C=9.$$