An airplane is flying at an airspeed of 690 km/hr in a cross-wind that is blowing from the northeast at a speed of 55 km/hr.
In what direction should the plane head to end up going due east? Give your answer correct to 2 decimal places.
I know that the wind is coming from NE, so it should have components w = -55cos(45) - 55sin(45)
However, I cannot for the life of me figure out the airplane's components.
Drat, shouldn't have erased the other.
Think of the ground as a map of the ground, and the path of the plane as a line that gets drawn on the map.
Only, turns out that the line is actually drawn on a transparent, plastic over-sheet - which is, like, the air.
With no wind, this makes no difference. But now imagine the transparent sheet being dragged while the line is being drawn. The starting point of the plane in the plastic (call it A) makes a path at 45 degrees (towards south west) while the line that gets drawn in the plastic is rather less than 45 degrees to the horizontal, and ends up with its left end at where A ends up, and its right-hand end dead horizontally level with where A started.
So you have a triangle of paths all of whose lengths are proportional to the speeds they got drawn, so you can do trigonometry to settle the unspecified speeds and angles.
Hope this helps.
Notice the triangle is in a similar configuration to the (horizontial aspect of the) other problem, but there you moved the plastic the opposite way (NE) along the 45 degree side, the horizontal side was the line drawn in the plastic, and problem was to find the length of the third, oblique side. Here, the 45 side is again the path of A (the starting point), while the oblique side is the line in the plastic, whose length you also know. You have to find the angle between these sides that puts their upper ends horizontally level.
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Originally Posted by MechyV 
