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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Heights and Distances

JEE Main · Mathematics · Heights and Distances

A vertical tower stands on a horizontal plane. Two points A and B...

A vertical tower stands on a horizontal plane. Two points AA and BB are located on the plane. The angles of elevation of the top of the tower from AA and BB are 30∘30^\circ and 45∘45^\circ, respectively. If the angle subtended by the line segment ABAB at the foot of the tower is 150∘150^\circ and the distance ABAB is 777\sqrt{7} meters, then the height of the tower (in meters) is

Options

  1. A77
  2. B777\sqrt{7}
  3. C737\sqrt{3}
  4. D772\dfrac{7\sqrt{7}}{2}

Correct answer

A. 77

Step-by-step solution

Let the height of the tower be hh and its foot be OO.

The distance of point AA from the foot of the tower is OA=hcot⁡30∘=h3OA = h \cot 30^\circ = h\sqrt{3}.

The distance of point BB from the foot of the tower is OB=hcot⁡45∘=hOB = h \cot 45^\circ = h.

In △AOB\triangle AOB, the angle subtended by ABAB at OO is ∠AOB=150∘\angle AOB = 150^\circ.

Applying the Cosine Rule in △AOB\triangle AOB:
AB2=OA2+OB2−2(OA)(OB)cos⁡150∘AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos 150^\circ

Substitute the known values:
(77)2=(h3)2+h2−2(h3)(h)(−32)(7\sqrt{7})^2 = (h\sqrt{3})^2 + h^2 - 2(h\sqrt{3})(h)\left(-\dfrac{\sqrt{3}}{2}\right)
343=3h2+h2+3h2343 = 3h^2 + h^2 + 3h^2
343=7h2343 = 7h^2
h2=49h^2 = 49
h=7h = 7

Thus, the height of the tower is 77 meters.

Answer: 77
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