JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Heights and Distances

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

4 5^{\circ}

. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of

3 0^{\circ}

to the horizontal plane, the angle of elevation of the top of the hill becomes

7 5^{\circ}

. Then the height of the hill (in metres) is ______.

Answer: 80

Solution

Let the hill have height

h

. From the starting point on the plane the elevation is

4 5^{\circ}

, so the horizontal distance to the foot equals

h

. Walking

80 m

up a

3 0^{\circ}

slope moves the observer

80 cos 3 0^{\circ} = 403 m

horizontally towards the hill and

80 sin 3 0^{\circ} = 40 m

upward. From there the remaining horizontal distance is

h - 403

and the remaining height is

h - 40

, and the new elevation is

7 5^{\circ}

with

tan 7 5^{\circ} = 2 + 3

, so

2 + 3 = \frac{h - 40}{h - 40 3}

. Expanding,

(2 + 3) (h - 403) = h - 40

reduces to

(1 + 3) h = 80 (3 + 1)

, giving

h = 80 m

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer