Question

If in $△ABC$, $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2$, then ${\sin }^{2}A:{\sin }^{2}B:{\sin }^{2}C$ can be

• A. $8:9:5$
• B. $8:5:9$
• C. $5:8:9$
• D. $5:8:5$

Answer 1

Answer: (B), (C)

The correct options are
B $8:5:9$
C $5:8:9$

As $A+B+C=\pi$, so
$\tan A+\tan B+\tan C=\tan A\tan B\tan C\Rightarrow 6=2\tan C\Rightarrow \tan C=3$
So, $\angle C$ lies in the first quadrant
$\sin C=\frac{3}{\sqrt{1}0}{\sin }^{2}C=\frac{9}{10}\cdots \left(1\right)$
Now,
$\tan A+\tan B=3\tan A\tan B=2\Rightarrow \tan A+\frac{2}{\tan A}=3\Rightarrow {\tan }^{2}A-3\tan A+2=0\Rightarrow (\tan A-1)(\tan A-2)=0\Rightarrow \tan A=1,2\Rightarrow \tan B=2,1$
So, $\angle A,\angle B$ are acute angles,
$\sin A=\frac{1}{\sqrt{2}},\frac{2}{\sqrt{5}}\Rightarrow {\sin }^{2}A=\frac{1}{2},\frac{4}{5}\Rightarrow {\sin }^{2}A=\frac{5}{10},\frac{8}{10}\cdots \left(2\right)\Rightarrow {\sin }^{2}B=\frac{8}{10},\frac{5}{10}\cdots \left(3\right)$
From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
${\sin }^{2}A:{\sin }^{2}B:{\sin }^{2}C=8:5:9\text{or}5:8:9$