If the angles of elevation of the top of a tower from three collinear points A ,   B and C on a line…

Let, the height of the tower$=h$

Let, $AB=x, BC=y, CD=z$

In $∆ADE,$$\tan {30}^o= \frac{h}{x+y+z}$

$\Rightarrow x+y+z=\sqrt{3}h ...\left(i\right)$

In $∆BDE, \tan 45°=\frac{h}{y+z}$

$\Rightarrow y+z=h ...\left(ii\right)$

In $∆CDE, \tan 60°=\frac{h}{z}$

$\Rightarrow z=\frac{h}{\sqrt{3}} ...\left(iii\right)$

From $\left(ii\right) \& \left(iii\right),$

$y=h\left(1- \frac{1}{\sqrt{3}}\right) ...\left(iv\right)$

From $\left(i\right) \& \left(ii\right),$

$x=\left(\sqrt{3}-1\right)h ...\left(v\right)$

$\therefore \frac{AB}{BC}=\frac{x}{y}= \frac{\left(\sqrt{3}-1\right)h}{h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)}$$= \frac{\sqrt{3}}{1}$