A tower $T_1$ of height $60 \mathrm{~m}$ is located exactly opposite to a tower $T_2$ of height $80 \mathrm{~m}$ on a straight road. From the top of $T_1$, if the angle of depression of the foot of $T_2$ is twice the angle of elevation of the top of $T_2$, then the width (in $m$ ) of the road…

Question

A tower $T_1$ of height $60 \mathrm{~m}$ is located exactly opposite to a tower $T_2$ of height $80 \mathrm{~m}$ on a straight road. From the top of $T_1$, if the angle of depression of the foot of $T_2$ is twice the angle of elevation of the top of $T_2$, then the width (in $m$ ) of the road between the feet of the towers $T_1$ and $T_2$ is

Aicharya · Accepted Answer

$ 	ext { Let the distance between } T_1 	ext { and } T_2 	ext { be } x $ From the figure $E A=60 \mathrm{~m}\left(T_1ight)$ and $D B=80 \mathrm{~m}\left(T_2ight)$ $\angle D E C=	heta$ and $\angle B E C=2 	heta$ Now in $	riangle D E C$, $ 	an 	heta=\frac{D C}{A B}=\frac{20}{x} $ and in $	riangle B E C$, $ 	an 2 	heta=\frac{B C}{C E}=\frac{60}{x} $ We know that $ 	an 2 	heta=\frac{2 	an 	heta}{1-(	an 	heta)^2} $ $ \begin{aligned} &\Rightarrow \frac{60}{x}=\frac{2\left(\frac{20}{x}ight)}{1-\left(\frac{20}{x}ight)^2} \ &\Rightarrow x^2=1200 \Rightarrow x=20 \sqrt{3} \end{aligned} $

A tower $T_1$ of height $60 \mathrm{~m}$ is located exactly opposite to a tower $T_2$ of height $80…

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