Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5. From a point on the plane, the angles of elevation of the bottom and top of the flagstaff are respectively 30° and 60°. Show that the height of the tower is 2.5 m.

Answer 1

Let $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that $BC=5$ m.
∠$CDA={30}^o$ and ∠$BDA={60}^o$
Let:
$AC=h$ m and $AD=x$ m

In right ∆​$CDA$, we have:
$\frac{AC}{AD}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x}=\frac{1}{\sqrt{3}}$

⇒ $x=h\sqrt{3}$

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

⇒ $\frac{(h+5)}{x}=\sqrt{3}$

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

⇒ $h+5=3h$
⇒ $2h=5$
⇒ $h=\frac{5}{2}=2.5$ m
∴ ​Height of the tower = $AC=h=2.5$ m