Given the figure:
We know that AB = 4, so we draw the height from F to AB, and we call it h:
From the figure, we know that:
[tex]\begin{gathered} \tan \theta=\frac{BE}{AB}=\frac{1}{4}=\frac{h}{AM}\Rightarrow AM=4h \\ \tan \alpha=\frac{AD}{AB}=\frac{2}{4}=\frac{h}{MB}\Rightarrow MB=2h \end{gathered}[/tex]
From this, we can say that:
[tex]\begin{gathered} 4h+2h=4 \\ h=\frac{2}{3} \\ \Rightarrow FN=\frac{4}{3} \\ \Rightarrow EN=1-h=1-\frac{2}{3}=\frac{1}{3} \\ \Rightarrow CN=1+EN=\frac{4}{3} \end{gathered}[/tex]
Then:
[tex]\tan x=\frac{FN}{CN}=\frac{4/3}{4/3}=1[/tex]
We conclude that x must be 45° because tan(45°) = 1.
