The angle of elevation of the top of a tower from a point A due north of it is $\alpha$ and from a point B at a distance of 9 units due west of $A$ is $\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is 15 units, then $\cot \alpha$ is equal to :

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Heights and Distances

The angle of elevation of the top of a tower from a point A due north of it is

α

and from a point B at a distance of 9 units due west of

A

cos^{- 1} (\frac{3}{13})

. If the distance of the point

B

from the tower is 15 units, then

cot α

is equal to :

(A) $\frac{6}{5}$ ✓ Correct
(B) $\frac{9}{5}$
(C) $\frac{4}{3}$
(D) $\frac{7}{3}$

Answer: (A)

Solution

Let O be the base of the tower with height h.

Given:

O B = 15

and

cos β = \frac{3}{13}

where

β

is the angle of elevation from B.

From

cos β = \frac{3}{13}

, we get

tan β = \frac{2}{3}

.

Since

tan β = \frac{h}{15}

h = 15 \times \frac{2}{3} = 10

In the horizontal plane,

O A^{2} + A B^{2} = O B^{2}

O A^{2} + 81 = 225

O A = 12

Therefore:

cot α = \frac{O A}{h} = \frac{12}{10} = \frac{6}{5}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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