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QJEEM20Mabb47: JEE Main 2020 Mathematics - Class 12 JEEM20Mabb47 Solution: The angle of elevation of the top of a hill from a point on the …

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45o. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30o to the horizontal plane, the angle of elevation of the top of the hill becomes 75o. Then the height of the hill (in meters) is ____________.

Step-by-Step Verified Solution

JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Height and Distance Question 21 English Explanation

sin 30o = x80{x \over {80}} ⇒\Rightarrow x = 40

cos 30o = y80{y \over {80}} ⇒\Rightarrow y = 40340\sqrt 3

Now, In Δ\Delta AEF

tan 75o = h−xh−y{{h - x} \over {h - y}}

⇒\Rightarrow 2 + 3\sqrt 3 = h−40h−403{{h - 40} \over {h - 40\sqrt 3 }}

⇒\Rightarrow (2+3)(h−403)\left( {2 + \sqrt 3 } \right)\left( {h - 40\sqrt 3 } \right) = h - 40

⇒\Rightarrow 2h - 80 3{\sqrt 3 } + 3{\sqrt 3 } h - 120 = h - 40

⇒\Rightarrow h + 3{\sqrt 3 } h = 80 + 80 3{\sqrt 3 }

⇒\Rightarrow ( 3{\sqrt 3 } + 1)h = 80( 3{\sqrt 3 } + 1)

⇒\Rightarrow h = 80 m

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