From a point on the ground, the angle of elevation of an aeroplane flying at an altitude 500m changes from 45q to 30q in 5 seconds. Find the speed of the aeroplane (in kmph). (a) 720(root 3- 1) (b) 720( root 3+ 1) (c) 360( root3- 1) (d) 360( root3+ 1)
Answers
Answer:
AD=3000m=3km [ given ]
⇒ AD=CB=3km
⇒ In △PAD,tan45
o
=
AP
AD
⇒ 1=
AP
3
⇒ AP=3km
⇒ In △PBC,tan30
o
=
PB
BC
⇒
3
1
=
PB
3
∴ PB=3
3
km
⇒ Now, AB=PB−AP
⇒ AB=3
3
−3=3(
3
−1)km
∴ AD=CD=3(
3
−1)km
⇒ Time taken to move from C to D=15sec=
60×60
15
=
3600
15
hours
⇒ So, Speed=
Time
Distance
=
Time
CD
⇒ Speed=
3600
15
3(
3
−1)
⇒ Speed=
15
3(
3
−1)×3600
⇒ Speed=720(
3
−1)=527.076km/hr (Take
3
=1.73)
Answer:
what does q stand for? is it degree , if yes then give a like to my answer so as to indicate and then i will proceed
ok leave it if u don't want to reply
i assume q as degree.
now see the attachment
tan45°=DE/AE
tan45°=0.5/AE
tan45°=1=0.5/AE
AE=0.5 ______________________i
now tan 30°=BC/AB
tan30°=BC/(AE+BE)
from (i) tan30°=0.5/(0.5+BE)
tan30°=1/√3=0.5/(0.5+BE)
1/√3=0.5/(0.5+BE)
0.5√3=0.5+BE
BE=0.5√3-0.5
BE=0.5(√3-1) __________________ii
now see my drawing attached to the answer
BE is the distance traveled in 5seconds
so, We know that S=VT(you must have studied it in std 9th , BTW, it's a GK)
BE=Vx5/3600 (I converted seconds to hour)
from (ii), 0.5(√3-1)=Vx1/720
=>V=360(√3-1)
Answer
