Skip to main content
Back to Height and Distance
JEE Main 2020
Height and Distance
Height and Distance
Easy

Question

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 45 o . After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30 o to the horizontal plane, the angle of elevation of the top of the hill becomes 75 o . Then the height of the hill (in meters) is ____________.

Answer: 30

Solution

sin 30 o = x80{x \over {80}} ⇒\Rightarrow x = 40 cos 30 o = y80{y \over {80}} ⇒\Rightarrow y = 40340\sqrt 3 Now, In Δ\Delta AEF tan 75 o = 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 - 803{\sqrt 3 } + 3{\sqrt 3 }h - 120 = h - 40 ⇒\Rightarrow h + 3{\sqrt 3 }h = 80 + 803{\sqrt 3 } ⇒\Rightarrow (3{\sqrt 3 } + 1)h = 80(3{\sqrt 3 } + 1) ⇒\Rightarrow h = 80 m

Previous QuestionNext Question

Practice More Height and Distance Questions

View All Questions