Question

In a triangle $ABC,$ if $\tan A:\tan B:\tan C=1:2:3$ then $a^2:b^2:c^2=$

• A. $5:8:9$
• B. $5:8:12$
• C. $3:5:8$
• D. $5:8:10$

Answer 1

Answer: (A)

$5:8:9$

Given: $\tan A:\tan B:\tan C=1:2:3$

$\Rightarrow \frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}=k$

$\Rightarrow \sum \tan A=\prod \tan A$

$\Rightarrow 6k=6k^3$

$\Rightarrow k(1-k^2)=0$

$\Rightarrow k\ne 0$ and $k^2=1$

$\Rightarrow k=1$

$\therefore \tan A=1,\tan B=2,\tan C=3$

$\Rightarrow \frac{1}{\cot A}=1,\frac{1}{\cot B}=2,\frac{1}{\cot C}=3$

or $\cot A=1,\cot B=\frac{1}{2},\cot C=\frac{1}{3}$

$\Rightarrow {\cot }^{2}A=1,{\cot }^{2}B=\frac{1}{4},{\cot }^{2}C=\frac{1}{9}$

Add $1$ to both sides,we get

$\Rightarrow 1+{\cot }^{2}A=2,1+{\cot }^{2}B=1+\frac{1}{4}=\frac{5}{4},1+{\cot }^{2}C=1+\frac{1}{9}=\frac{10}{9}$

We know that ${\csc }^2\theta =1+{\cot }^2\theta$

$\Rightarrow {\csc }^{2}A=2$

${\csc }^{2}B=\frac{5}{4}$

${\csc }^{2}C=\frac{10}{9}$

$\Rightarrow {\sin }^{2}A=\frac{1}{2},{\sin }^{2}B=\frac{4}{5},{\sin }^{2}C=\frac{9}{10}$

Using sine rule, we have

$\Rightarrow a^2:b^2:c^2=\frac{1}{2}:\frac{4}{5}:\frac{9}{10}$

$=5:8:9$ (on simplification)