Question

If in triangle $ABC,\frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}$, then

• A. $6\sqrt{2}a=3\sqrt{5}b=2\sqrt{10}c$
• B. $3\sqrt{2}a=4\sqrt{5}b=7\sqrt{10}c$
• C. $6\sqrt{2}a=4\sqrt{5}b=2\sqrt{10}c$
• D. None of these

Answer 1

The correct option is A $6\sqrt{2}a=3\sqrt{5}b=2\sqrt{10}c$

We have, $\frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}=k$

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

Since $A,B,C$ are the angles of a triangle

$\therefore A+B+C=\pi$

$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\Rightarrow k+2k+3k=k.2k.3k$$\Rightarrow 6k=6k^3\Rightarrow k=1.$

$[$Hence, $k\ne 0$ and $k$ cannot be negative as that will make $\tan A,\tan B,\tan C$ all negative meaning thereby that angles $A,B,C$ are all $>{90}^0$ which is not possible in a triangle$].$

Hence $\tan A=1\Rightarrow \sin A=\frac{1}{\sqrt{2}};$$\tan B=2\Rightarrow \sin B=\frac{2}{\sqrt{5}}$

and, $\tan C=3\Rightarrow \sin C=\frac{3}{\sqrt{10}}$

Now, from sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow a\sqrt{2}=b.\frac{\sqrt{5}}{2}=\frac{c\sqrt{10}}{3}$

$\Rightarrow 6\sqrt{2}a=3\sqrt{5}b=2\sqrt{10}c.$