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QJEEM19Ma1f29: JEE Main 2019 Mathematics - Class 12 JEEM19Ma1f29 Solution: The angle of elevation of the top of a vertical tower standing…

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45o from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30o, then the distance (in m) of the foot of the tower from the point A is :

Multiple Choice Options

  • A 15(1+3)15\left( {1 + \sqrt 3 } \right)
  • B 15(3−3)15\left( {3 - \sqrt 3 } \right)
  • C 15(3+3)15\left( {3 + \sqrt 3 } \right)
  • D 15(5−3)15\left( {5 - \sqrt 3 } \right)

Step-by-Step Verified Solution

JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Height and Distance Question 24 English Explanation xd=tan⁡30o⇒xd=13⇒d=3x{x \over d} = \tan {30^o} \Rightarrow {x \over d} = {1 \over {\sqrt 3 }} \Rightarrow d = \sqrt 3 x

and x+30d=tan⁡45o{{x + 30} \over d} = \tan {45^o} ⇒\Rightarrow d = x + 30

d=d3+30⇒(1−13)d=30d = {d \over {\sqrt 3 }} + 30 \Rightarrow \left( {1 - {1 \over {\sqrt 3 }}} \right)d = 30

⇒d=3033−1⇒d=303(3+1)2=153(3+1)\Rightarrow d = {{30\sqrt 3 } \over {\sqrt 3 - 1}} \Rightarrow d = {{30\sqrt 3 (\sqrt 3 + 1)} \over 2} = 15\sqrt 3 (\sqrt 3 + 1)

⇒\Rightarrow d = 15 ( 3 + 3\sqrt 3 )

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