an aeroplane flying horizontally 1200 meter above the ground is observed at an angle of elevation of 60°. after 6 second it's angle of elevation is observed to be 30°. find the speed of the aeroplane in km/hr.
Answers
Answer:
480√3 km/hr
Step-by-step explanation:
Let E be the initial position of aeroplane and A be its position after 6 sec.
AB = EC = 1200 m
tan 30 = AB/ED = 1200/BD = 1/√3
BD = 1200√3 m
tan 60 = EC/CD = 1200/CD = √3
CD = 1200/√3 m
AE = BC = BD - CD = 1200√3 - 1200/√3
AE = 1200 (√3 - 1/√3)
AE = 1200(2√3/3)
AE = 800√3 m
when time, t = 6 sec
distance, d = AE = 800√3 m
Speed = distance/time
= 800√3/6
= 400√3m / 3 sec
= (400√3 / 1000 × 3600/3) km/hr
= 480√3 km/hr
Step-by-step explanation:
In angle BAC=60° and angle DAB=30°
In ∆ABC
tan 60°=opposite side/adjacent side
√3=2500/AC. {cross multiplication}
AC=2500/√3
In ∆AED
tan 30°=opposite side/adjacent side
1/√3 =2500/AE. {cross multiplication}
AE=2500√3
BD=CE
CE = AE-AC
BD=2500√3-2500/√3
BD=2500(√3-1/√3)
BD=2500(√3×√3)/√3
BD=2500(3-1)/√3
BD=2500(2)/√3
BD=5000/√3
in kilometres 100 meters =1km
BD=5/√3
Hence aeroplane travelled 5/√3 in 15 seconds
Time= 15 seconds (given)
Time=15/3600
= 1/240 hrs
speed = distance/time
speed=5/√3÷1/240 {cross multiplication}
speed= 5×240/1.732 {√3=1.732}
=1200/1.732
=692.840647 km/HR
hence the speed of the aeroplane=692.840647 km/hr
