A tower T_1 of height 60 , m is located exact | vedclass

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(D) Let the distance between $T_1$ and $T_2$ be $x$.From the figure,$EA = 60 \, m$ $(T_1)$ and $DB = 80 \, m$ $(T_2)$.Let $C$ be a point on $T_2$ such

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$20\sqrt{3}$

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(D) Let the distance between $T_1$ and $T_2$ be $x$.From the figure,$EA = 60 \, m$ $(T_1)$ and $DB = 80 \, m$ $(T_2)$.Let $C$ be a point on $T_2$ such that $EC$ is horizontal. Then $EC = AB = x$.$DC = DB - CB = 80 - 60 = 20 \, m$.Given $\angle DEC = 	heta$ (angle of elevation of top of $T_2$) and $\angle BEC = 2	heta$ (angle of depression of foot of $T_2$).In $\Delta DEC$,$	an 	heta = \frac{DC}{EC} = \frac{20}{x}$.In $\Delta BEC$,$	an 2	heta = \frac{BC}{EC} = \frac{60}{x}$.Using the identity $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,we have:$\frac{60}{x} = \frac{2(\frac{20}{x})}{1 - (\frac{20}{x})^2}$.$\frac{60}{x} = \frac{40/x}{1 - 400/x^2} = \frac{40x}{x^2 - 400}$.$60(x^2 - 400) = 40x^2$.$60x^2 - 24000 = 40x^2$.$20x^2 = 24000$.$x^2 = 1200$.$x = \sqrt{1200} = 20\sqrt{3} \, m$.

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