Question

In a $\Delta ABC$, let $\tan A=1,\tan B=2,\tan C=3$ and $c=3$. Then the area (in sq. units) of the $\Delta ABC$ is:

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

Answer 1

The correct option is B $3$

Given: $\tan A=1,\tan B=2,\tan C=3$ and $c=3$
$\tan A=1\Rightarrow \sin A=\frac{1}{\sqrt{2}}$
$\tan B=2\Rightarrow \sin B=\frac{2}{\sqrt{5}}$
$\tan C=3\Rightarrow \sin C=\frac{3}{\sqrt{10}}$
Now, using sine law, $a\sqrt{2}=\frac{b\sqrt{5}}{2}=\frac{c\sqrt{10}}{3}$
$\therefore a\sqrt{2}=\frac{b\sqrt{5}}{2}=\sqrt{10}(\because c=3)$
So, we have $a=\sqrt{5};b=2\sqrt{2};c=3$
$\therefore$ Area $=\frac{1}{2}a\cdot b\cdot \sin C=\frac{1}{2}\sqrt{5}\cdot \left(2\sqrt{2}\right)\frac{3}{\sqrt{10}}=3$