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QJEEM18Ma692e: JEE Main 2018 Mathematics - Class 12 JEEM18Ma692e Solution: A man on the top of a vertical tower observes a car moving at a …

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

Multiple Choice Options

  • A 9(1+3)9\left( {1 + \sqrt 3 } \right)
  • B 18(1+3)18\left( {1 + \sqrt 3 } \right)
  • C 18(3−1)18\left( {\sqrt 3 - 1} \right)
  • D 92(3−1){9 \over 2}\left( {\sqrt 3 - 1} \right)

Step-by-Step Verified Solution

JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Height and Distance Question 31 English Explanation

Assume height of tower = AB = h

From Δ\Delta ABD,

tan45o = ABAD{{AB} \over {AD}}

⇒\Rightarrow    \,\,\, ABAD=1{{AB} \over {AD}} = 1 [ as    tan45o = 1]

⇒\Rightarrow    \,\,\, AB = AD

∴   \therefore\,\,\, AD = h

From Δ\Delta BAC,

tan30o = ABAC{{AB} \over {AC}}

⇒\Rightarrow    \,\,\, hAC=13{h \over {AC}} = {1 \over {\sqrt 3 }}

⇒\Rightarrow    \,\,\, AC = h 3\sqrt 3

As,    AC = AD + DC

⇒\Rightarrow    \,\,\, DC = AC −- AD = 3h\sqrt 3 h −- h

Given that, time taken to reach from point C to D = 18 min.

∴   \therefore\,\,\, Car speed = distancetime{{distance} \over {time}} = CD18{{CD} \over {18}} = (3−1)h18{{(\sqrt 3 - 1)h} \over {18}}

∴   \therefore\,\,\, Time taken to move from D to A

= Distance  of  DAspeed{{Distan ce\,\,of\,\,DA} \over {speed}}

= h(3−1)h18{h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}

= 18(3−1){{18} \over {\left( {\sqrt 3 - 1} \right)}}

= 18(3+1)3−1{{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}

= 9 (3+1)\left( {\sqrt 3 + 1} \right) min.

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