Question

The length of the string between a kite and a point on the ground is 90 meters. If the string makes angle theta with the ground level such that tan theta is equal to 15 by 8 how high is the kite flying assuming that there is no slack in the string.

Answer 1

$tan\theta =\frac{15}{8}=\frac{BC}{AB}=\frac{h}{AB}\Rightarrow h=ABtan\theta \dots \dots \left(1\right)$
Now, in $\Delta ABC$
$(AC{)}^2=(BC{)}^2+(AB{)}^2$ [By Pythagoras]
$\Rightarrow (90{)}^2=h^2+(AB{)}^2\Rightarrow AB=\sqrt{(90{)}^2-h^2}$
Put in (1)
$\Rightarrow h=\left(\sqrt{(90{)}^2-h^2}\right)tan\theta$ Squaring both side
$\Rightarrow h^2=\left(\right(90{)}^2-h^2)ta{n}^2\theta \Rightarrow h^2=(90{)}^2+ta{n}^2\theta -h^{2}ta{n}^2\theta \Rightarrow h^2(1+ta{n}^2\theta )=(90{)}^{2}ta{n}^2\theta$
Now put value of $tan\theta =\frac{15}{8}$
$\Rightarrow h^2=\frac{8100\times 225}{289}\Rightarrow h=\sqrt{\frac{8100\times 225}{289}}h=79.4m$