Question

In a triangle $ABC$ if $\tan A=\frac{1}{2},\tan B=k+\frac{1}{2}$ and $\tan C=2k+\frac{1}{2}$, then the possible value of $\left[k\right]$, where [.] represents greatest integer function is

Answer 1

Given $\tan A=\frac{1}{2},\tan B=k+\frac{1}{2}$ and $\tan C=2k+\frac{1}{2}$
In $△ABC$
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\Rightarrow 3k+\frac{3}{2}=\frac{1}{2}(k+\frac{1}{2})(2k+\frac{1}{2})$
$\Rightarrow 8k^2-18k-11=0$
$\Rightarrow k=\frac{44}{16},k=-\frac{1}{2}$
Here, $k=-\frac{1}{2}$ (not possible as $\tan B=-\frac{1}{2}+\frac{1}{2}=0$)
Hence, possible value of k is $\frac{44}{16}=2.75$
So, $\left[k\right]=2$