Question

aeroplane flying at a height of 4000 m from the ground passes vertically above another plane at the instant when the angle between the two places 45 degree respectively find the vertical distance between 2 friends ​

Answer 1

Answer:

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Step-by-step explanation:

Let the height of first plane be AB=4000m and the height of second plane be BC=xm

∠BDC=45

∘

and ∠ADB=60

∘

In △CBD

(i)

y

x

=tan45

∘

=1

⇒x=y

In △ABD,

y

4000

=tan60

∘

=

3

⇒x=y=

3

4000

3

=2306.67m

∴ the vertical distance between two=4000−y=4000−2306.67=1693.33m

solution