The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle θ with the level ground such that tan θ = $\Big(\dfrac{15}{8}\Big)$, how high is the kite?

Let the required height AB = h metres and length of a string AC = 85 m.

Given,

tan θ = $\Big(\dfrac{15}{8}\Big) = \dfrac{\text{perpendicular}}{\text{base}}$

Let the Height be 15x and Base be 8x.

Hypotenuse = $\sqrt{(15x)^2 + (8x)^2}$

Hypotenuse = $\sqrt{289x^2}$ = 17x

In triangle ABC,

$\Rightarrow \sin \theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{h}{85} \\[1em] \Rightarrow \dfrac{15x}{17x} = \dfrac{h}{85} \\[1em] \Rightarrow \dfrac{15}{17} \times 85 = h \\[1em] \Rightarrow h = 5 \times 15 \\[1em] \Rightarrow h = 75 \text{ m}.$

Hence, height of kite from ground = 75 m.