Question

The angle of elevation of the top of a tree from a point A on the ground is 60°. On walking 20 metres away from its base, to a point B, the angle of elevation changes to 30°.Find the height of the tree.

Answer 1

Let $CD$ be the tree and $A$ be the point on the ground such that ∠$DAC\hspace{0.17em}={60}^o$, $AB=20$m and ∠$DBC\hspace{0.17em}={30}^o$.
Let:
$CD=h$ m and $AC\hspace{0.17em}=x$ m

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

⇒ $\frac{h}{x}=\sqrt{3}$ (or) $x=\frac{h}{\sqrt{3}}$
In the right ∆$DBC$, we have:
$\frac{CD}{BC}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{(x+20)}=\frac{1}{\sqrt{3}}$
⇒ $\sqrt{3}h=x+20$

On putting $x=\frac{h}{\sqrt{3}}$, we get:
$\sqrt{3}h=\frac{h}{\sqrt{3}}+20$
⇒ $3h=h+20\sqrt{3}$
⇒ $2h=20\sqrt{3}$
⇒ $h=\frac{20\sqrt{3}}{2}=10\sqrt{3}=17.32$ m
Hence, the height of the tree is 17.32 m.