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Question
An aeroplane when flying at a height 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 30° and 60° respectively. Find the distance between the two planes at that instant.
Solution
Let C and D be the two aeroplanes and A be the point of observation. Then,
∠CAB = 30°, ∠DAB = 60°, BC = 3125 m
Let DC = y m, AB = x m
In right ΔABC, ∠B = 90°
tan 30° = `("BC")/("AB")`
⇒ `1/sqrt(3) = 3125/("AB")`
⇒ AB = `3125sqrt(3)` m ...(i)
In right ΔABD, ∠B = 90°
tan 60° = `("BD")/("AB")`
⇒ `sqrt(3) = ("y"+ 3125)/(3125 sqrt(3))` ...[From equation (i)]
⇒ 3125 × 3 = y + 3125
⇒ y = 3125 (3 – 1)
⇒ y = 2 × 3125
⇒ y = 6250 m
Therefore, the distance between the two planes is 6250 m.
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