JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Heights and Distances

A vertical pole fixed to the horizontal ground is divided in the ratio

3 : 7

by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :

(A)
$810$
(B)
$1215$
(C)
$1210$
✓ Correct
(D)
$610$

Answer: (C)

Solution

Let the pole height be

h = 3 k + 7 k = 10 k

. The mark is at height

3 k

from ground. At distance 18 m, if both parts subtend equal angles

α

, then

tan α = \frac{3 k}{18}

and

tan (2 α) = \frac{10 k}{18}

. Using

tan (2 α) = \frac{2 tan α}{1 - tan ^{2} α}

\frac{10 k}{18} = \frac{2 \cdot \frac{3 k}{18}}{1 - ( \frac{3 k}{18} ) ^{2}}

. Simplifying:

\frac{10 k}{18} = \frac{6 k /18}{1 - 9 k ^{2} /324} = \frac{6 k \cdot 324}{18 ( 324 - 9 k ^{2} )} = \frac{108 k}{324 - 9 k ^{2}}

. Cross-multiplying:

10 k (324 - 9 k^{2}) = 18 \cdot 108 k

, giving

3240 - 90 k^{2} = 1944

, so

90 k^{2} = 1296

, thus

k^{2} = 14.4

k = \frac{6 10}{5}

. Height

h = 10 k = 1210

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Heights and Distances

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