Answer :

Let C be position of station 1 and D of station 2 and BC = x.

Given, CD = 4000 km

Now, in triangle ABC,

∠ACB = 60°

We know,

⇒ AB = √3BC

⇒ AB = x√3 ……… (1)

And, in triangle ABD,

∠ADB = 30°

We know,

……… (2)

Now, equating (1) & (2), we get–

⇒ 3 x = x + 4000

⇒ 3x – x = 4000

⇒ 2x = 4000

⇒ x = 2000

And distance between the satellite and earth(AB) = x√3

= 2000(1.732)

= 3464 km

∴ Distance between the satellite and earth = 3464 km

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