How to find $\cot (x)$ of this square?

Question

So according to question, what values of $\cot (x)$? So I've tried and got $\frac{-1}{4}$ Thanks for helping!

Answer 1

Produce $AE$ and $BC$ to meet at $K$. Then $\triangle KEC\sim\triangle KAB$. As $EC:AB=2:6=1:3$, we have $KC:KB=1:3$ and hence $KC=3$. So $\displaystyle \tan\angle KEC=\frac{2}{3}$ and $\displaystyle \tan\angle KBF=\tan\angle AFB=\frac{6}{3}=2$. Note that $x=\angle KEC+\angle KBF$.

\begin{align*} \tan x&=\frac{\tan\angle KEC+\tan\angle KBF}{1-\tan\angle KEC\tan\angle KBF}\\ &=\frac{\frac{2}{3}+2}{1-(\frac{2}{3})(2)}\\ &=-8 \end{align*}

$\displaystyle \cot x=\frac{-1}{8}$.