A tower T_1 of height 60 ~m is located exactly opposite to a tower T_2 of height 80 ~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

Correct answer: D. 20 3 Let the distance between T_1 and T_2 be x From the figure E A=60 m (T_1) and D B=80 m(T_2) D E C=θ and B E C=2 θ Now in D E C, θ=(D C)/(A B)=(20)/(x) and in B E C, 2 θ=(B C)/(C E)=(60)/(x) We know that 2 θ=(2 θ)/(1-( θ)^2) aligned & (60)/(x)=(2((20)/(x)))/(1-((20)/(x))^2) & x²=1200 x=20 3 aligned

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Mathematics
Chapter: Heights and Distances

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Heights and Distances

A tower T_1 of height 60 ~m is located exactly opposite to a tower...

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

A tower T_1 of height 60 ~m is located exactly opposite to a tower T_2 of height 80 ~m on a straight road. From the top

Options

A $20 \sqrt{2}$
B $10 \sqrt{2}$
C $10 \sqrt{3}$
D $20 \sqrt{3}$

Correct answer

D. $20 \sqrt{3}$

Step-by-step solution

\text{ Let the distance between }

T_1

\text{ and }

T_2

\text{ be }

<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252FsvbdBQle0gWepePzNmIBW6KL31B0ksE4P033i896-uM.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FsvbdBQle0gWepePzNmIBW6KL31B0ksE4P033i896-uM.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_main%2FsvbdBQle0gWepePzNmIBW6KL31B0ksE4P033i896-uM.original.fullsize.png?alt=media"> \\ From the figure

E A=60

\mathrm{ m} \left(T_1\right)

and

D B=80 \mathrm{ m}\left(T_2\right)\angle D E

C=\theta and \angle B E C=2 \theta Now in \triangle D E C, \tan \theta=\frac{D C}{A B}=\frac{20}{x}

and in \triangle B E C, \tan 2 \theta=\frac{B C}{C E}=\frac{60}{x}

We know that \tan 2 \theta=\frac{2 \tan \theta}{1-(\tan \theta)^2}

\begin{aligned}&\Rightarrow \frac{60}{x}=\frac{2\left(\frac{20}{x}\right)}{1-\left(\frac{20}{x}\right)^2}\\ &\Rightarrow x²=1200 \Rightarrow x=20 \sqrt{3} \end{aligned}

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