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QJEEM18Mae678: JEE Main 2018 Mathematics - Class 12 JEEM18Mae678 Solution: A tower T 1 of height 60 m is located exactly opposite to a…

A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is :

Multiple Choice Options

  • A 10210\sqrt 2
  • B 10310\sqrt 3
  • C 20320\sqrt 3
  • D 20220\sqrt 2

Step-by-Step Verified Solution

Let the distance between T1 and T2 be x

JEE Main 2018 (Online) 15th April Evening Slot Mathematics - Height and Distance Question 32 English Explanation

From the figure

EA = 60 m (T1) and DB = 80 m (T2)

∠DEC=θ\angle DEC = \theta and ∠BEC=2θ\angle BEC = 2\theta

Now in ∠DEC\angle DEC ,

tan⁡θ=DCAB=20x\tan \theta = {{DC} \over {AB}} = {{20} \over x}

and in ΔBEC\Delta BEC ,

tan⁡2θ=BCCE=60x\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}

We know that

tan⁡2θ=2tan⁡θ1−(tan⁡θ)2\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}

⇒\Rightarrow    60x=2(20x)1−(20x)2{{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}

⇒\Rightarrow    x2=1200{x^2} = 1200   ⇒\Rightarrow   x=203x = 20\sqrt 3

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