Question

The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is

(a) $50\sqrt{3}\mathrm{m}$
(b) $100\sqrt{3}\mathrm{m}$
(c) $50\sqrt{2}\mathrm{m}$
(d) 100 m

Answer 1

(a) $50\sqrt{3}\mathrm{m}$
Let $AB$ be the string of the kite and $AX$ be the horizontal line.
If $BC$ ⊥ $AX$, then $AB=100$ m and ∠$BAC={60}^o$.
Let:
$BC=h$m

In the right ∆$ACB$, we have:
$\frac{BC}{AB}=\sin {60}^o=\frac{\sqrt{3}}{2}$

⇒ $\frac{h}{100}=\frac{\sqrt{3}}{2}$
⇒ $h=\frac{100\sqrt{3}}{2}=50\sqrt{3}$ m
Hence, the height of the kite is $50\sqrt{3}$ m.