JEE Main · 2018 · 08 Apr 2018 · Offline · Heights and Distances

PQR is a triangular park with

PQ = PR = 200 m

. A T.V. tower stands at the mid-point of QR . If the angles of elevation of the top of the tower at P, Q and R are respectively

4 5^{\circ}, 3 0^{\circ}

and

3 0^{\circ}

, then the height of the tower (in m ) is :

(A) $502$
(B)100✓ Correct
(C)50
(D) $1003$

Answer: (B)

Solution

Let the tower of height

h

stand at

T

, the midpoint of

QR

. The

4 5^{\circ}

elevation from

P

gives

P T = h

, and the

3 0^{\circ}

elevation from

Q

gives

QT = h 3

. Since

P Q = P R = 200

and

T

is the midpoint of

QR

P T

is the altitude of the isosceles triangle and

P T ⊥ QT

, so

P Q^{2} = P T^{2} + Q T^{2}

gives

20 0^{2} = h^{2} + 3 h^{2} = 4 h^{2}

. Hence

h^{2} = 10000

and

h = 100

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

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