Question

A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point O on the ground is 60° and the angle of depression of the point O from the top of the tower is 45°. Show that the height of the tower is $\frac{5}{2}(\sqrt{3}+1)\mathrm{m}.$

Answer 1

Let $AB$ be the tower and $BC$ be the pole such that $BC=5$ m.
∠​$BDA={45}^o$ and ∠$CDA={60}^o$
Let:
$AB=h$ m and $AD=x$ m

In right ∆​$BAD,$ we have:
$\frac{AB}{AD}=\tan {45}^o=1$

⇒ $\frac{h}{x}=1$
⇒ x = h
Or,
$h=x$

In right ∆$CAD$, we have:
$\frac{AC}{AD}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{(h+5)}{x}=\sqrt{3}$

Putting the value of $x=h$ in the above equation, we have:
$\frac{(h+5)}{h}=\sqrt{3}$

⇒ $h+5=h\sqrt{3}$
⇒ $h(\sqrt{3}-1)=5$
⇒ $h=\frac{5}{(\sqrt{3}-1)}=\frac{5}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{5}{2}(\sqrt{3}+1)$ m

∴ Height of the tower = $AB=h=\frac{5}{2}(\sqrt{3}+1)$ m