Question

angle of elevation of an aeroplane from a point on ground is 60 degree. after a flight of 30 seconds angle of elevation becomes 30 degree if aeroplane is flying on constant height of 3000√3metre,find speed of plane

Answer 1

First find the base of 30° angle formed
√⅓ = 3000√3/B1
Base 1= 9000 m

Now take the 60° triangle that formed and find the base of that
√3 Base 2=3000√3
Base 2 = 3000m

Now the distance between Base 1 and base 2
9000m - 3000m = 6000 m

speed = 6000 m/ 30s
200m/s

Ans= 200m/s

Hope it help you Dear

Answer 2

Answer:

Step-by-step explanation:IN ∆ ADE

TAN30°=DE /AD

1/√3=300/AD

300√3=AD

IN∆ABC

TAN60°=BC/AD+DC

√3=300/300√3+DC

√3(300√3+DC)=300

300×3+√3DC=300

900+√3DC=300

√3DC=600

DC=600/√3

THEREFORE

DC=BE

30SEC = 30 SEC

30 = 600/√3

30 =200√3

30√3=200

=200-30√3

10(20-3√3)

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