JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Heights and Distances

The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be

4 5^{\circ}

from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be

3 0^{\circ}

, then the distance (in m) of the foot of the tower from the point A is:

(A) $15 (1 + 3)$
(B) $15 (3 - 3)$
(C) $15 (3 + 3)$ ✓ Correct
(D) $15 (5 - 3)$

Answer: (C)

Solution

Let the tower height be

h

and the foot be a horizontal distance

d

from

A

. From

A

the

4 5^{\circ}

elevation gives

h = d

. From

B

, which is

30

m above

A

, the

3 0^{\circ}

elevation gives

\frac{1}{3} = \frac{h - 30}{d}

. Putting

h = d

d - 30 = \frac{d}{3}

, so

d (1 - \frac{1}{3}) = 30

, giving

d = \frac{30 3}{3 - 1} = 15 (3 + 3)

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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