Question

In a triangle ABC, prove that $\sin A\cdot \sin B\cdot \sin C=\frac{1}{8}(3+\sqrt{3})$ and $\cos A\cdot \cos B\cos C=\frac{1}{8}(\sqrt{3}-1)$. Find the angles of $\Delta$ ABC.

Answer 1

Dividing the given relations, we get
$\tan A\cdot \tan B\cdot \tan C=\frac{\sqrt{3})\sqrt{3}+1)}{(\sqrt{3}-1)}=\frac{\sqrt{3}(\sqrt{3}+1{)}^2}{2}$
or $\tan A\cdot \tan B\cdot \tan C=\sqrt{3}(2+\sqrt{3})=\sum \tan A$
$S_1=S_3=\sqrt{3}(2+\sqrt{3})$
Now we have to find the value of
$S_2=\sum \tan A\cdot \tan B$
$\cos (A+B+C)=\cos A\cdot \cos B\cdot \cos C(1-\sum \tan A\cdot \tan B)$
$-1=\frac{1}{8}(\sqrt{3}-1)[1-S_2]$
or $S_2=1+\frac{8}{\sqrt{3}-1}$
$\therefore S_2=1+4(\sqrt{3}+1)=5+4\sqrt{3}$
Hence $\tan A,\tan B,\tan C$ are the roots of
$x^3-x^2{S}_1=x{S}_2-S_3=0$
or $x^3-\sqrt{3}(2+\sqrt{3})x^2+(5+4\sqrt{3})x-\sqrt{3}(2+\sqrt{3})=0$
Clearly $x=1$ satisfies above.
$\therefore (x-1)[x^2-2(\sqrt{3}+1)x+\sqrt{3}(2+\sqrt{3}]=0$
or $(x-1)(x-\sqrt{3})[x-(2+\sqrt{3}\left)\right]=0$
$\therefore x=1,\sqrt{3},2+\sqrt{3}$
$\therefore \tan \theta =\tan {45}^o,\tan {60}^o,\tan {75}^o$
$\therefore A={45}^o,B={60}^o,C={75}^o$.