Question 583736
An airplane with a speed of 160 knots is headed east while a 24-knot wind is blowing from 240. Find the course and round to the nearest degree.
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The vectors make a triangle with 160 on the base, a 150 deg angle and the 24 side.  If you plot it on the x-y plane with the plane's speed from (0,0) to (160,0) and the wind vector starting at (160,0) length 24, the internal angle is 150 degs.
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find the 3rd side:
{{{r = sqrt(160^2 + 24^2 - 2*160*24*cos(150))}}}
r =~ 181.18 knots (the plane's groundspeed)
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Use the Sine Law to find the angle at the origin:
181.18/sin(150) = 24/sin(A)
sin(A) = 24*sin(150)/181.18
A =~ 3.8 degs --> 4 degs
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The course = 090 - 4 = 086
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Note:  East is 090 degs heading, and the angles increase clockwise
The plane is flying a heading of 090 @ 160 knots
Winds are 240 @ 24 kts.
"Winds are two-four-zero at two-four." is what is said.