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Home » An aeroplane when flying at a heigt of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 300 and 600 respectively. Find the distance between the two planes at the instant
An aeroplane when flying at a heigt of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 300 and 600 respectively. Find the distance between the two planes at the instant
ML Agarwal Heights and distances class 10 Maths
An aeroplane when flying at a heigt of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 300 and 600 respectively. Find the distance between the two planes at the instant

Solution:

Consider the distance between two planes = h m

It is given that

AD = 3125 m, ∠ACB = 600 and ∠ACD = 300

In triangle ACD

tan 300 = AD/AC

Substituting the values

1/√3 = 3125/AC

AC = 3125√3 ……. (1)

In triangle ABC

tan 600 = AB/AC

Substituting the values

√3 = (AD + DB)/ AC

So we get

√3 = (3125 + h)/ AC

AC = (3125 + h)/ √3 ….. (2)

Using both the equations

(3125 + h)/ √3 = 3125√3

By further calculation

h = (3125√3 × √3) – 3125

h = 3125 × 3 – 3125

h = 9375 – 3125

h = 6250 m

Therefore, the distance between two planes at the instant is 6250 m.

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