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We are given that a kite is flying at \[120m\] height and the string is out \[130m\]. Therefore a figure can be made

Here BC is the height of the kite. So \[BC = 120m\] and we suppose \[AC = ym\] and \[BA = x\] metre

Therefore, we can apply Pythagoras theorem in the above right angled triangle

\[ \Rightarrow {y^2} = {x^2} + {(120)^2} - - - - - - - (i)\]

Now it is given that kite is moving away horizontally at the rate of \[52{\text{ }}m/s\]

Which means \[\dfrac{{dx}}{{dt}} = 52m/s - - - - - - - - (ii)\]

Where \[\dfrac{{dx}}{{dt}}\] is rate of horizontal movement of kite which is equal to \[52{\text{ }}m/s\]

Since \[\dfrac{{dx}}{{dt}}\] is given, One another \[\dfrac{{dx}}{{dt}}\] can also be found by differentiating \[(i)\] with respect to time. So differentiating 1 w.r.t ‘\[t\]’, we get

\[ \Rightarrow 2y\dfrac{{dy}}{{dt}} = 2x\dfrac{{dx}}{{dt}} + 0 - - - - - - - - (iii)\]

Where \[\dfrac{d}{{dt}}({x^2}) = 2x\dfrac{{dx}}{{dt}},\dfrac{d}{{dt}}({y^2}) = 2y\dfrac{{dy}}{{dt}}\]and \[\dfrac{d}{{dt}}{(120)^2} = 0\]because derivative of constant is zero and \[\dfrac{d}{{dt}}({x^n}) = n{x^{n - 1}}\dfrac{{dx}}{{dt}}\]

From \[(iii)\]we get

\[ \Rightarrow 2y\dfrac{{dy}}{{dt}} = 2x\dfrac{{dx}}{{dt}}\]

\[2\] get cancel on both sides

\[ \Rightarrow \]\[y\dfrac{{dy}}{{dt}} = x\dfrac{{dx}}{{dt}}\]

\[ \Rightarrow \]\[\dfrac{{dy}}{{dt}} = \dfrac{x}{y}\dfrac{{dx}}{{dt}}\]

From \[(ii)\] putting the value \[\dfrac{{dx}}{{dt}}\] = \[52{\text{ }}m/s\]. Here and we get

\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{x}{y}(52)\]

Also given \[y{\text{ }} = {\text{ }}130m\] and \[x{\text{ }} = {\text{ }}50m\]

Putting here the values of \[x\] and \[y\], we get

\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{50}}{{130}}(52)\]

On solving, we get

\[ \Rightarrow \dfrac{{dy}}{{dt}} = 20m/s\]

Therefore, string is paid at the rate of \[20m/s\] where \[x{\text{ }} = {\text{ }}50\] was found by using Pythagoras theorem in given right angled triangle

Which means ,

\[ \Rightarrow A{C^2} = A{B^2} + B{C^2}\]

\[ \Rightarrow {(130)^2} = {(AB)^2} + {(120)^2}\]

\[ \Rightarrow 16900 = {(AB)^2} + 14400\]

From her, \[AB{\text{ }} = {\text{ }}x\] mitre

\[ \Rightarrow 16900{\text{ }} = {\text{ }}{x^2} + 14400\]

From the above equation, value of \[x\] can be found

\[ \Rightarrow 16900{\text{ }} = {\text{ }}{x^2} + 14400\]

On taking square root, we get

\[ \Rightarrow {x^2} = 16900 - 14400\]

\[ \Rightarrow {x^2} = 2500\]

\[x = 50m\]

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