Multiple choice

An airplane when flying at a height of 5000 m from the ground passes vertically above another airplane at an instant, when the angles of elevation of the two airplane from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. The vertical distance between the airplane at that instant is:

  1. $5000 (\sqrt{3}-1)m$
  2. $5000 (3-\sqrt{3})m$
  3. $5000 (1-\frac{1}{\sqrt{3}})m$

  4. 4500 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the point on the ground be at distance x from the vertical line. Height of top plane = 5000. tan(60) = 5000/x => x = 5000/sqrt(3). Let height of lower plane be h. tan(45) = h/x => h = x = 5000/sqrt(3). Vertical distance = 5000 - 5000/sqrt(3) = 5000(1 - 1/sqrt(3)).

AI explanation

The horizontal distance from the observation point to the planes is found using the lower plane by the tangent ratio, where tan(45 degrees) equals 5000 divided by the distance, making the distance 5000 m. Let the vertical distance between the planes be x; the height of the higher plane from the ground is (5000 minus x), so tan(60 degrees) equals (5000 minus x) divided by 5000. Substituting the square root of 3 for tan(60 degrees) gives 5000 multiplied by the square root of 3 equals 5000 minus x, so x equals 5000 multiplied by (1 minus 1 divided by the square root of 3) m.

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