The angle of elevation of an aeroplane from point A on the ground is 60°.After a flight of 30 seconds,the angle elevation changes to 30°.If the plane is flying at a constant height of 3600√3 m,
find the speed in km/hr of the plane.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)
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Let C & E be the initial position of the plane above the ground and E the position of plane after 30 seconds.
Given ∠CAB = 60°, ∠EAD = 30°, BC = DE = 3600√3 m
In ΔABC,
tan 60° = BC/AB
√3 = 3600√3/AB
AB= 3600√3/√3
AB = 3600 m
In ΔADE,
tan 30° = CE/AD
1/√3= 3600√3/AD
AD= 3600×(√3×√3)
AD= 3600×3
AD= 10800 m
CE= BD = AD - AB
CE= 10800 – 3600 = 7200 m
The plane Travels 7200 m in 30 seconds.
Speed of plane = distance/ time
Speed of plane= 7200 / 30= 720/3=240 m/sec
Speed of plane=( 240×60×60)/1000 km/h
= 864 km/h
Hence, speed of plane= 864 km/h.
HOPE THIS WILL HELP YOU....
Given ∠CAB = 60°, ∠EAD = 30°, BC = DE = 3600√3 m
In ΔABC,
tan 60° = BC/AB
√3 = 3600√3/AB
AB= 3600√3/√3
AB = 3600 m
In ΔADE,
tan 30° = CE/AD
1/√3= 3600√3/AD
AD= 3600×(√3×√3)
AD= 3600×3
AD= 10800 m
CE= BD = AD - AB
CE= 10800 – 3600 = 7200 m
The plane Travels 7200 m in 30 seconds.
Speed of plane = distance/ time
Speed of plane= 7200 / 30= 720/3=240 m/sec
Speed of plane=( 240×60×60)/1000 km/h
= 864 km/h
Hence, speed of plane= 864 km/h.
HOPE THIS WILL HELP YOU....
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