JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Heights and Distances

The angle of elevation of the summit of a mountain from a point on the ground is

4 5^{\circ}

. After climbing up one km towards the summit at an inclination of

3 0^{\circ}

from the ground, the angle of elevation of the summit is found to be

6 0^{\circ}

. Then the height (in km ) of the summit from the ground is:

Answer: (C)

Solution

Let

h

be the height of the summit. From the ground point,

tan 4 5^{\circ} = \frac{h}{x + y}

gives

h = x + y

. Climbing

1 km

3 0^{\circ}

raises the observer by

z = sin 3 0^{\circ} = \frac{1}{2}

and moves them horizontally by

y = cos 3 0^{\circ} = \frac{3}{2}

. From the new position

tan 6 0^{\circ} = \frac{h - z}{x}

gives

3 x = h - \frac{1}{2}

, and with

x = h - y = h - \frac{3}{2}

this becomes

3 (h - \frac{3}{2}) = h - \frac{1}{2}

, which simplifies to

(3 - 1) h = 1

, so

h = \frac{1}{3 - 1} km

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option