an aeroplane is flying at a height of 3000m an observer find the angle of elevation to be 60 degree and after 6sec it changes to 45degree find the speed of the aeroplane
Answers
Answer:
222.145 m/s
Step-by-step explanation:
This question is on trigonometry.
The two right angled triangles formed.
They have a common height h = 3000 meters.
The height is opposite to both the angles of elevation.
We will therefore use Tan.
The distance on the ground between the observer and the plane at the different angles of elevation is given by :
Tan 60 = 3000/x
x = 3000/Tan 60
= 3000/1.3764 = 2179.60 meters
Tan 45 = 3000/y
y = 3000/Tan 45
= 3000/0.8541 = 3512.47
The difference between the two distances gives the distance traveled by the plane in 6 seconds.
= 3512.47 - 2179.60 = 1332.87 m
Speed = Distance / time = 1332.87/6 =
= 222.145 m/s
Answer:
S = 366 m/s
Step-by-step explanation:
The question is one of an Application of Trigonometry,
In the bigger triangle where the angle of elevation 45 degree let the base assumed to be "D",
In the smaller triangle where the angle of elevation 60 degree let the base assumed to be "d",
now,
D= 3000/tan45
D= 3000 m
d= 3000/tan60
d= 3000/\sqrt{3} m
The distance traveled by plane,
D-d
= 3000 - 3000/\sqrt{3}
= 3000 (1 - \sqrt{3})
= 3000(0.732)
= 2196 meters
Then comes the basic distance speed time relationship formula,
S= D/T
= 2196/6
= 366 m/s