Question

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

Answer 1

Given : $A+B+C=\pi$
We know that,
$\tan A+\tan B+\tan C=\tan A\tan B\tan C\Rightarrow \frac{3}{2}+3k=\frac{(2k+1)(4k+1)}{8}\Rightarrow \frac{3(2k+1)}{2}=\frac{8k^2+6k+1}{8}\Rightarrow 12(2k+1)=(2k+1)(4k+1)\Rightarrow (2k+1)[4k+1-12]=0\Rightarrow (2k+1)(4k-11)=0\Rightarrow k=-\frac{1}{2},\frac{11}{4}$

Checking the values of $k$, we get
When $k=-\frac{1}{2}$, we get
$\tan B=0\Rightarrow B=n\pi$
This is not possible for any angle of triangle
When $k=\frac{11}{4}$, we get
$\tan B=\frac{13}{4},\tan C=6$

Hence, $k=\frac{11}{4}\Rightarrow \left[k\right]=2$