Question

achei por aqui uma resolução dessa mssma questão, mas se puder ser um pouquinho mais esclarecedor, ficarei grata :)

Answer 1

$\begin{matrix}\tan30^\circ=\dfrac{h}{240-x}\\h=(240-x).\tan30^\circ\end{matrix}\qquad\qquad\qquad\begin{matrix}\tan45^\circ=\dfrac{h}{x}\\h=x.\tan45^\circ\end{matrix}\\\\\\(240-x).\tan30^\circ=x.\tan45^\circ\\(240-x).\dfrac{\sqrt3}{3}=x.1\\\dfrac{240.\sqrt3}{3}-\dfrac{x.\sqrt3}{3}=x\\80.\sqrt3=x+\dfrac{x.\sqrt3}{3}\\80.\sqrt3=x.\left(1+\dfrac{\sqrt3}{3}\right)\\80.\sqrt3=x.\left(\dfrac{3+\sqrt3}{3}\right)\\x=\dfrac{240.\sqrt3}{3+\sqrt3}\\x=\dfrac{240.\sqrt3}{3+\sqrt3}.\dfrac{3-\sqrt3}{3-\sqrt3}\\x=\dfrac{240.(3.\sqrt3-3)}{9-3}\\x=\dfrac{240.(3.(\sqrt3-1)}{6}\\x=40.(3.(\sqrt3-1))\\\boxed{x=120.(\sqrt3-1)\,\text{m}}$