Question

The length of a string between a kite and a point on the ground is $90$m if the string makes an angle $\theta$ with the ground level such that $\tan \theta =\frac{15}{8}$, how high is the kite?Assume that there is no slack in the string.

Answer 1

$tan\theta =\frac{15}{8}=\frac{BC}{AB}=\frac{h}{AB}$

$\Rightarrow h=ABtan\theta$ ______ (1)

By Pythagoras Theorem,

${90}^2=h^2+A{B}^2$

$=A{B}^{2}ta{n}^2\theta +A{B}^2$

$\Rightarrow A{B}^2(1+ta{n}^2\theta )={90}^2$

$\Rightarrow A{B}^2=\frac{{90}^2}{1+ta{n}^2\theta }=\frac{{90}^2}{1+\frac{{15}^2}{8^2}}=\frac{{90}^2\times 64}{289}$

$\Rightarrow AB=\sqrt{\frac{{90}^2\times 64}{289}}=\frac{90\times 8}{17}=42.35m$

$\therefore AB=42.35m$

$h=ABtan\theta =42.35\times \frac{15}{8}=\frac{90\times 8}{17}\times \frac{15}{8}$

$=\frac{90\times 15}{17}=79.42m$

Hence height of the kite $=79.42m$