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) $100 \mathrm{~m}$
Solution:
(a) $50 \sqrt{3} \mathrm{~m}$
Let $A B$ be the string of the kite and $A X$ be the horizontal line.
If $B C \perp A X$, then $A B=100 \mathrm{~m}$ and $\angle B A C=60^{\circ}$.
Let:
$B C=h \mathrm{~m}$
In the right $\triangle A C B$, we have:
$\frac{B C}{A B}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{h}{100}=\frac{\sqrt{3}}{2}$
$\Rightarrow h=\frac{100 \sqrt{3}}{2}=50 \sqrt{3} \mathrm{~m}$
Hence, the height of the kite is $50 \sqrt{3} \mathrm{~m}$.