If a plane flying 60 degrees northeast at 200 km/hr is struck by a wind that blows due west at 50 km/hr, what angle must the plane fly to maintain the sixty degrees northeast bearing.
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To be honest, I really have no idea how to start this.
If a plane flying 60 degrees northeast at 200 km/hr is struck by a wind that blows due west at 50 km/hr, what angle must the plane fly to maintain the sixty degrees northeast bearing.
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To be honest, I really have no idea how to start this.
1st: make a sketch
Identify all that is given.
After 1 hour where would the the plane be if there were no problem with the wind?
(Call the starting or origin Point A and call the point where the airplane would be after 1 hour as Point B.)
You need to determine that.
Hint: at a speed of 200km/h with a heading of 60degrees northeast,
determine the km distance NORTH of the starting point
& the km distance EAST of the starting point.
[ N = 200 cos(60); E = 200 sin(60) ]
(The wind (at 50 km/h) will effectively move the airplane 50 km WEST in 1 hour. This is what you need to counteract or correct.)
THEN, from that location where the plane would be after 1 hour (which was point B), move it directly EAST 50 km (call this point C).
From the starting point (A), the airplane needs to head towards point C, to correct for the wind.
The adjusted bearing is:
(You've already computed the values for N & E above.)
$\displaystyle arc\tan\left(\dfrac{E+50}{N}\right)$
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Notice that this is NOT a question about "wind resistance" as if, say wind were pushing a sailboat. The airplane is actually supported by the wind just as if you were rolling a toy car over a table while carrying the table to one side. Add the two speeds "vectorially".
One thing you can do is separate into "components". Since cos(60)= $\displaystyle \sqrt{3}/2$, the eastward component of the planes flight, relative to the wind, is $\displaystyle (200)\sqrt{3}/2= 100\sqrt{3}$. Similarly, since sin(60)= 1/2, the northward component is (1/2)(200)= 100.
The wind is blowing directly west so its eastward component is -50 while its northward component is 0. Adding those gives a vector whose eastward component is $\displaystyle \100\sqrt{3}- 50$ and whose northward component is $\displaystyle 100$. You can represent that as a right triangle having legs of length $\displaystyle 100\sqrt{3}- 50$ and 100. Thinking of the north component as "opposite side" and the east component as "near side" $\displaystyle tan(\theta)= \frac{100\sqrt{3}- 50}{100}$ where $\displaystyle \theta$ is the angle the direction makes "north of east" or simply "northeast".