A kite is flying at a height of 75 m from th | vedclass

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(B) Let the height of the kite from the ground be $AB = 75 \ m$.Let the angle made by the string with the ground be $\angle ACB = \Theta$.Given that $

Vedclass · Accepted Answer

$85$

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(B) Let the height of the kite from the ground be $AB = 75 \ m$.Let the angle made by the string with the ground be $\angle ACB = \Theta$.Given that $\cot \Theta = 8/15$.In the right-angled triangle $	riangle ABC$,$\cot \Theta = \frac{	ext{Base}}{	ext{Perpendicular}} = \frac{BC}{AB}$.Therefore,$\frac{BC}{75} = \frac{8}{15}$.$BC = \frac{8 	imes 75}{15} = 8 	imes 5 = 40 \ m$.The length of the string is the hypotenuse $AC$.Using the Pythagorean theorem,$AC = \sqrt{AB^2 + BC^2}$.$AC = \sqrt{75^2 + 40^2} = \sqrt{5625 + 1600} = \sqrt{7225}$.$AC = 85 \ m$.

$A$ kite is flying at a height of $75 \ m$ from the ground. The string makes an angle $\Theta$ (where $\cot \Theta = 8/15$) with the level ground. Assuming that there is no slack in the string,the length of the string is equal to (in $m$):

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