Q67 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
In a $\triangle A B C$, if $\tan A: \tan B: \tan C=1: 2: 3$ and $\sin A: \sin B: \sin C=\sqrt{5}: 2 \sqrt{2}: k$, then $k=$
- A$\sqrt{2}$
- B2
- C$\sqrt{3}$
- Correct answerD3
Solution
Given, $\tan A: \tan B: \tan C=1: 2: 3$
$\begin{aligned}
& \tan A=\frac{\sin A}{\cos A}=\frac{a / 2 R}{\frac{b^2+c^2-a^2}{2 b c}}=\left(\frac{1}{b^2+c^2-a^2}\right) \frac{2 a b c}{2 R} \\
& \therefore \tan A: \tan B: \tan C=\frac{1}{b^2+c^2-a^2} \\
& \frac{1}{a^2+c^2-b^2} \frac{1}{a^2+b^2-c^2}
\end{aligned}$
Clearly, $\tan A: \tan B: \tan C=l: m: n$
Then, $a: b: c=\sqrt{\frac{1}{m}+\frac{1}{n}}: \sqrt{\frac{1}{n}+\frac{1}{l}}: \sqrt{\frac{1}{l}+\frac{1}{m}}$
Here, $l: m: n=1: 2: 3$
$\begin{aligned}
& a: b: c=\sqrt{\frac{1}{2}+\frac{1}{3}}: \sqrt{\frac{1}{3}+1}: \sqrt{1+\frac{1}{2}} \\
& a: b: c=\frac{\sqrt{5}}{\sqrt{6}}: \frac{2 \sqrt{2}}{\sqrt{6}}: \frac{3}{\sqrt{6}}
\end{aligned}$
Given, $\sin A: \sin B: \sin C=\sqrt{5}: 2 \sqrt{2}: k \quad \therefore \quad k=3$
Hence, the correct answer is option: 'D'.