Aeroplane Problem!

Geometry Level 2

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


The answer is 2100.

Ayush G Rai by Ayush G Rai , 20, India

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1 solution

In Δ A B D ⟹ tan ⁡ 6 0 ∘ = 3150 x ⟹ x = 3150 3 In Δ A B C ⟹ tan ⁡ 3 0 ∘ = a x ⟹ a = 3150 3 = 1050 Distance between two planes = 3150 − 1050 = 2100 . \large\displaystyle \text{In } \Delta ABD\\ \large \displaystyle \implies \tan 60^{\circ} = \frac{3150}{x} \implies x=\frac{3150}{\sqrt3}\\ \large \displaystyle \text{In } \Delta ABC\\ \large\displaystyle \implies \tan 30^{\circ} = \frac{a}{x} \implies a = \frac{3150}{3} = 1050\\ \large \displaystyle \text{Distance between two planes } = 3150 - 1050 = \color{#3D99F6}{\boxed{2100}}.

Details: tan ⁡ θ = Opposite Side Adjacent Side \large \displaystyle \text{Details: }\\ \large \displaystyle \tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}}

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Excellent solution with a good diagram..+1

Ayush G Rai - 5 years, 1 month ago

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Thank You! ⌣ ¨ \ddot\smile

Samara Simha Reddy - 5 years, 1 month ago

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