Multiple choice

An aeroplane when flying at a height 2500 m from the ground passes vertically above another aeroplane At an instant when the angles of elevation of the two aeroplanes from the same point on the ground are $\displaystyle 45^{\circ}$ and $\displaystyle 30^{\circ} $ respectively then the vertical distance between the two aeroplanes at that instant is

  1. 1158.5 m

  2. 1056.5 m

  3. 1008.5 m

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let distance from point to ground projection be x. Height of plane 1 = 2500. tan(45) = 2500/x -> x = 2500. Height of plane 2 = h. tan(30) = h/x -> h = 2500 * tan(30) = 2500 * 0.577 = 1443.5. Vertical distance = 2500 - 1443.5 = 1056.5 m.

AI explanation

Let the point on the ground be at a distance x meters from the point directly below the higher aeroplane. Using the tangent ratio for the higher aeroplane, tan 45 degrees equals 2500 divided by x, so x equals 2500. For the lower aeroplane, tan 30 degrees equals its height divided by 2500, giving the lower height as 2500 divided by root 3, which is approximately 1443.5 meters. Subtracting this from the higher altitude gives 2500 minus 1443.5, resulting in 1056.5 meters.

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