Question

If A, B, C are the angles of a triangle, show that "In a triangle ABC if $\tan A:\tan B:\tan C=3:4:5$ then the value of $\sin A\sin B\sin C$ is"?

• A. $\frac{2}{\sqrt{5}}$
• B. $\frac{2\sqrt{5}}{3}$
• C. $\frac{2\sqrt{5}}{9}$
• D. $\frac{2}{3\sqrt{5}}$

Answer 1

The correct option is B $\frac{2\sqrt{5}}{3}$

Let $\tan A=3\lambda ,\tan B=4\lambda ,$ and $\tan C=5\lambda$. In a triangle $S_1=S_3$
$\therefore (3+4+5)\lambda =\left(\mathrm{3.4.5}\right){\lambda }^3\therefore 12\lambda -60{\lambda }^3=0$
$\therefore \lambda =\pm \frac{1}{\sqrt{5}}$
$\therefore \tan A=\frac{3}{\sqrt{5}},\tan B=\frac{4}{\sqrt{5}},\tan C=\frac{5}{\sqrt{5}}$
$\therefore \sin A=\frac{3}{\sqrt{14}},\sin B=\frac{4}{\sqrt{21}},\sin C=\frac{5}{\sqrt{30}}$
$\therefore \sin A\sin B\sin C=\frac{\mathrm{3.4.5}}{7\sqrt{\mathrm{2.3.30}}}=\frac{60}{7.6\sqrt{5}}=\frac{2\sqrt{5}}{7}$
In a triangle, $\sin A,\sin B,\sin C$ are all $+$ive and hence we have chosen only $+$ sign out of $\pm$.