A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of  $22.5°\text{\hspace{0.17em}and\hspace{0.17em}}67.5°$.  What is the ratio of the height of the taller tower to the height of the shorter tower?  $(\therefore \tan 22.5°=\sqrt{2}-1)$

Let $ED$ be the taller tower and

$AB$ be the shorter tower

Let  $C$ be the point of observation

Given that  $\angle ACB=22.5°$ and  $\angle DCE=67.5°$

Given that  $C$ is the midpoint of  $BD$

$\therefore BC=CD$

In  $\Delta ABC\Rightarrow \tan 22.5°=\frac{AB}{BC}\to \left(1\right)$

In $\Delta CDE\Rightarrow \tan 67.5°=\frac{DE}{BCD}\to \left(2\right)$

From (1) and (2)

$\frac{\tan 67.5°}{\tan 22.5°}=\frac{\frac{DE}{CD}}{\frac{AB}{BC}}(\therefore BC=CD)$

$\Rightarrow \frac{\tan (90-22.5)}{\tan 22.5}=\frac{DE}{AB}$

$\Rightarrow \frac{\cot 22.5°}{\tan 22.5°}=\frac{ED}{AB}$

$\Rightarrow \frac{1}{{(\tan 22.5°)}^2}=\frac{DE}{AB}$

$\Rightarrow \frac{1}{{(\sqrt{2}-1)}^2}=\frac{DE}{AB}$

$\Rightarrow {\left[\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}\right]}^2=\frac{DE}{AB}$

$\Rightarrow \frac{{(\sqrt{2}+1)}^2}{{(2-1)}^2}=\frac{DE}{AB}$

$\Rightarrow 2+1+2\sqrt{2}=\frac{DE}{AB}$

$\Rightarrow \frac{3+2\sqrt{2}}{1}=\frac{DE}{AB}$

$\therefore DE:AB=3+2\sqrt{2}:1$

$\therefore$ Required ratio $=DE:AB$

$=(3+2\sqrt{2}):1$