The length of a string between a kite and a point on the ground is 90 metres. If the string makes an angle θ with the ground level such that tan θ = 15/8, how high is the kite? Assume that there is no slack in the string.
Let be the kite of height m and the length of string between kite and a point on the ground is meters. Someters and
Now we have to find the height of kite.
We have the corresponding figure as follows
Given that $\tan \theta=\frac{15}{8}$
Since $\sec ^{2} \theta=1+\tan ^{2} \theta$
$\Rightarrow \quad \sec ^{2} \theta=1+\frac{225}{64}$
$\Rightarrow \quad \sec ^{2} \theta=\frac{289}{64}$
$\Rightarrow \quad \cos ^{2} \theta=\frac{64}{289}$
$\Rightarrow \quad \cos \theta=\frac{8}{17}$
Again,
Since $\sin ^{2} \theta=1-\cos ^{2} \theta$
$\Rightarrow \quad \sin ^{2} \theta=1-\frac{64}{289}$
$\Rightarrow \quad \sin ^{2} \theta=\frac{289-64}{289}$
$\Rightarrow \quad \sin \theta=\frac{15}{17}$
Now, in a triangle
$\Rightarrow \quad \sin \theta=\frac{h}{90}$
$\Rightarrow \quad \frac{15}{17}=\frac{h}{90}$
$\Rightarrow \quad h=\frac{15 \times 90}{17}$
$\Rightarrow \quad h=79.41$
Hence the height kite is $79.41$ meters.