Help with a Vector Problem

[krypt0nite]

An airplane, whose air speed is 480 km/h is suppose to fly in a straight path 28.0 degrees N of E. But a steady 100k/m wind is blowing from the north. In what direction should the plane head.

Well my problem is I'm having problem drawing this vector. I can't seem to get that 28degrees N of E inside the angle of my vector diagram. Someone please show me how to draw it.

[HallsofIvy]

An airplane, whose air speed is 480 km/h is suppose to fly in a straight path 28.0 degrees N of E. But a steady 100k/m wind is blowing from the north. In what direction should the plane head.

Well my problem is I'm having problem drawing this vector. I can't seem to get that 28degrees N of E inside the angle of my vector diagram. Someone please show me how to draw it.


Uh, that is the easy part of the problem! Draw a horizontal line representing "East" at the left end it measure an angle up 28 degrees. Draw an arrow from the left end of your horizontal line in that direction. Draw an arrow straight down representing the wind (which I hope is actually 100 km/hr) with "length" 100. Connect your beginning and end points to form a triangle. That last line should have "length" 480 and its direction is the direction in which the plane should head.

[krypt0nite]

Could you draw that in paint or something really quickly because its hard to understand. But anyways the answer i got is 40 degrees N of S is the direction it should head in. Is that correct?

[HallsofIvy]

No, I can't use Paint: the file is much too large to upload.

Surely you don't mean "N of S"! If you got 40 degrees N of E, that's pretty close.

In my picture, the angle between the horizontal axis and the "flight path" is 28 degrees. By "alternate interior angles" the angle the horizontal makes with that same "flight path" at the other end is also 28 degrees and so the angle inside the triangle at that point (from the "flight path" to north) is 90+ 28= 118 degrees. Now use the sine law: that 118 degree angle is opposite the 480 km/h line. The angle opposite the 100 km/h line is given by sin(A)/100= sin(118)/480 so A is about 10.6 degrees. The heading of the airplane should be that plus the original 28: about 38.6 degees North of East.