plz help me friends
Answers
speed of plane = 200km/h
》200×5/18 = 500/9 m/s
》distance = 500/9 × 54
》= 500×6 = 3000m = 3km
》Here , AC = 3KM.
》angle B= 20 degree.
》Height (BC )= ?
》sin20 = BC/AC
》0.342 = BC/ 3
》BC = 3×0.342
》(BC = 1.026 km) ans
■I HOPE this helps
Nice question.
The responsible figure as an example is attached with the answer.
In the figure, the plane is at C. And the airport is at A.
The length of AB is equal to the height of the plane from the ground when it started landing.
The plane from C reaches the airport A through AC.
Here we have to find AC.
Given that the speed of the plane is 200 km/h.
Convert this speed into m/s by multiplying it by 5/18.
200 × 5/18
⇒ 1000/18 m/s
⇒ 500/9 m/s
So, the speed of plane is 500/9 m/s.
The plane reached the ground after 54 seconds.
When we multiply this 54 by the speed of the plane in m/s, we get the length of AC.
⇒ 500/9 × 54
⇒ 3000 m
So, AC = 3000 m (or we can say 3 km. But I'm taking 3000 m.)
Consider ΔABC.
sin 20 = AB / AC
⇒ 0.342 = AB / 3000
⇒ AB = 0.342 × 3000
⇒ AB = 1026 m
⇒ AB = 1.026 km
So the answer is 1026 m, or we can say 1.026 km.
Hope this helps you.
Plz ask me if you have any doubt on my answer, and please don't report my answer.
Thank you. :-))
