Question

An aeroplane, when flying at an height of 4000m from the ground passes vertically above another aeroplane
at an instant when the angle of elevation of the two planes from the same point on the ground are 45 and 60
respectively. Find the vertical distance between both the aeroplanes. (takreV3 = 1.73)​

Answer 1

Answer:

• Vertical distance between both the aeroplanes is 1688 m.

Step-by-step explanation:

Diagram:

• Refers to attachment.

Given:

• Height of 1st plane = 4000 m

• Angle of elevation of two plane from same point on the ground are 45° and 60°.

To Find:

• Vertical distance between two points.

Let Vertical distance between two planes be 'h'.

Now, In ΔACD

$\longrightarrow \sf \tan 60^{\circ}=\dfrac{AC}{CD}\\ \\ \\ \longrightarrow \sf \sqrt{3}=\dfrac{4000}{x}\\ \\ \\ \sf Now,\;in\;\triangle BCD,\\ \\ \\ \longrightarrow \sf  \tan 45^{\circ}=\dfrac{BC}{CD}\\ \\ \\ \longrightarrow \sf 1=\dfrac{4000-h}{x}\\ \\ \\ \longrightarrow \sf x=4000-h\\ \\ \\ \longrightarrow \sf \dfrac{4000}{\sqrt{3}}   =4000-h\\ \\ \\ \longrightarrow \sf h=4000-\dfrac{4000}{\sqrt{3}}\\ \\ \\ \longrightarrow \sf h= \dfrac{6920-4000}{1.73}\\ \\ \\ \longrightarrow \sf h=\dfrac{2920}{1.73}$

$\large{\boxed{\longrightarrow {\sf h=1688\;m}}}$

Hence, Vertical distance between both the aeroplanes is 1688 m.

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