Multiple choice

From the top of a tower, the angle of depression of a point on the ground is $60^o$ . If the distance of this point from the tower is $ \dfrac {1}{\sqrt{3}+1}$ m, then the height of the tower is :

  1. $ \dfrac {4 \sqrt{3}}{2}$ m
  2. $ \dfrac { \sqrt{3} + 3}{2}$ m
  3. $ \dfrac {3 - \sqrt{3}}{2}$ m
  4. $ \dfrac { \sqrt{3} }{2}$ m
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C Correct answer
Explanation

The height h = distance * tan(60). Given distance = 1 / (sqrt(3) + 1), h = (1 / (sqrt(3) + 1)) * sqrt(3). Rationalizing the denominator: sqrt(3) * (sqrt(3) - 1) / (3 - 1) = (3 - sqrt(3)) / 2.

AI explanation

In a right triangle formed by the tower and the point on the ground, the angle of depression equals the angle of elevation from the point, which is 60 degrees. Using the basic trigonometric ratio for tangent, we have tan 60 degrees equals the height of the tower divided by the distance from the tower. Substituting the given distance gives the height equal to the square root of 3 multiplied by (1 divided by (the square root of 3 plus 1)), and rationalizing this expression by multiplying the numerator and denominator by (the square root of 3 minus 1) yields (3 minus the square root of 3) divided by 2 meters.

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