JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Heights and Distances

A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that

QR = 15

m. If from a point A on the ground the angle of elevation of R is

6 0^{\circ}

and the part PR of the tower subtends an angle of

1 5^{\circ}

at A, then the height of the tower is:

Answer: (A)

Solution

Let

P Q = h

be the height of the tower, with

QR = 15

m and

P R = x

m, so

h = 15 + x

. From point A on the ground, the angle of elevation to R is

6 0^{\circ}

, so

tan 6 0^{\circ} = \frac{15}{A Q}

, giving

A Q = \frac{15}{3} = 53

. The angle of elevation to P is

6 0^{\circ} + 1 5^{\circ} = 7 5^{\circ}

, so

tan 7 5^{\circ} = \frac{15 + x}{A Q}

. We have

tan 7 5^{\circ} = 2 + 3

, so

15 + x = 53 (2 + 3) = 103 + 15

, giving

x = 103

. Therefore,

h = 15 + 103 = 5 (3 + 23)

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option