Multiple choice

An aeroplane is flying horizontally at a height of 3150 m above a horizontal plane ground. At a particular instant it passes another plane vertically below it. At this instant, the angles of elevation of the planes from a point on the ground are $30^o$ and $60^o$.Hence, the distance between the two planes at that instant is :

  1. 1050 m

  2. 2100 m

  3. 4200 m

  4. 5250 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the point on the ground be O. The height of the plane is 3150. For the 60-degree angle, the base distance is 3150/tan(60) = 3150/sqrt(3) = 1050*sqrt(3). For the 30-degree angle, the base distance is 3150/tan(30) = 3150*sqrt(3). The distance between the planes is the difference in their heights if they were at the same horizontal distance, but here they are vertically aligned. The distance between the two planes is 3150 - (3150/3) = 2100m.

AI explanation

Let the horizontal distance from the observation point to the point directly below the planes be x. Using the tangent function for the higher plane, tan 60 degrees equals 3150 divided by x, which simplifies to the square root of 3 equals 3150 divided by x, giving x equals 1050 times the square root of 3. For the lower plane, tan 30 degrees equals its height divided by x, so its height is x divided by the square root of 3. Substituting x gives the lower plane's height as 1050 m, and subtracting this from the higher plane's altitude of 3150 m gives the distance between the two planes as 2100 m.

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