Height and Distance
2020
Q1
JEE Mains
Numerical
14 Mar 2026
Let AD and BC be two vertical poles
at A and B respectively on a horizontal ground.
If AD = 8 m, BC = 11 m and AB = 10 m; then the distance
(in meters) of a point M on AB from the point A such
that MD2 + MC2 is minimum is ______.
at A and B respectively on a horizontal ground.
If AD = 8 m, BC = 11 m and AB = 10 m; then the distance
(in meters) of a point M on AB from the point A such
that MD2 + MC2 is minimum is ______.
Correct Answer: 5
Explanation:
(MD)2 + (MC)2 = h2 + 64 + (h – 10)2 + 121
= 2h2 – 20h + 64 + 100 + 121
= 2(h2 – 10h) + 285
= 2(h – 5)2 + 235
It is minimum if h = 5.
2020
Q2
JEE Mains
Numerical
14 Mar 2026
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 ____________.
Correct Answer: 80
Explanation:
sin 30o = ${x \over {80}}$ $ \Rightarrow $ x = 40
cos 30o = ${y \over {80}}$ $ \Rightarrow $ y = $40\sqrt 3 $
Now, In $\Delta $AEF
tan 75o = ${{h - x} \over {h - y}}$
$ \Rightarrow $ 2 + $\sqrt 3 $ = ${{h - 40} \over {h - 40\sqrt 3 }}$
$ \Rightarrow $ $\left( {2 + \sqrt 3 } \right)\left( {h - 40\sqrt 3 } \right)$ = h - 40
$ \Rightarrow $ 2h - 80${\sqrt 3 }$ + ${\sqrt 3 }$h - 120 = h - 40
$ \Rightarrow $ h + ${\sqrt 3 }$h = 80 + 80${\sqrt 3 }$
$ \Rightarrow $ (${\sqrt 3 }$ + 1)h = 80(${\sqrt 3 }$ + 1)
$ \Rightarrow $ h = 80 m