JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Heights and Distances

ABC is a triangular park with

AB = AC = 100

metres. A vertical tower is situated at the mid-point of

B C

. If the angles of elevation of the top of the tower at

A

and

B

are

cot^{- 1} (32)

and

cosec^{- 1} (22)

respectively, then the height of the tower (in metres) is:

(A)25
(B) $105$
(C) $\frac{100}{3 3}$
(D)20✓ Correct

Answer: (D)

Solution

Let the tower of height

h

stand at

D

, the midpoint of

B C

. From

B

cosec β = 22

gives

sin β = \frac{1}{2 2}

, so

B D = 7 h

. From

A

cot α = 32

gives

A D = 32 h

. In the isosceles triangle with

A B = A C = 100

A D ⊥ B C

, so

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

becomes

10 0^{2} = 18 h^{2} + 7 h^{2} = 25 h^{2}

, giving

h^{2} = 400

and

h = 20

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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