Why Are Trigonometric Component Rules Reversed for Vectors?

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madinsane
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Homework Statement



An airplane starting from airport A flies 300 km east, then 350 km at 30.0" west of north, and then 150 km north to arrive finally at airport B. The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight?

I actually solved it... I drew the vectors and got the resultant displacement.
so its 300i
150j
and then we use rsinθ and rcosθ to get the x and y components of west of north. At first I used rcosθ to get x and rsinθ to get y but that turned out to be wrong (it was actually rsinθ that equaled to x and rcosθ equal to y)
That is what I don't get!
How come the rule here is reversed?
Thanks


The Attempt at a Solution

 
on Phys.org
madinsane said:

Homework Statement



An airplane starting from airport A flies 300 km east, then 350 km at 30.0" west of north, and then 150 km north to arrive finally at airport B. The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight?

I actually solved it... I drew the vectors and got the resultant displacement.
so its 300i
150j
and then we use rsinθ and rcosθ to get the x and y components of west of north. At first I used rcosθ to get x and rsinθ to get y but that turned out to be wrong (it was actually rsinθ that equaled to x and rcosθ equal to y)
That is what I don't get!
How come the rule here is reversed?
Thanks


The Attempt at a Solution


Perhaps you just mechanically use rsinθ that equaled to x and rcosθ equal to y when you are comparing to the positive x-axis - (or the only polar axis)