An airline route from San Francisco to Honolulu is on a bearing of 233 degrees. A jet flying at 450 mph on that bearing runs into a wind blowing at 39.0 mph from a direction 114.0 degree. Find the resulting bearing and groundspeed?

Question

anonymous · Answer

We have a vector of 450 mph at 233 mph, and
solving for the drawing triangle, we have an
interior angle of 233-180 = 53 degrees.

The x component is 450*cos(53) = 270.816760418 and
the y component is 450*sin(53) = 359.385979521.

Since this vector is in quadrant 3, the
coordinates would be (-270.82, -359.39).

Since the wind is blowing *from* 114 degrees, it
is blow *toward* 114+180 = 294 degrees. To solve
for the drawn triangle, we can see that the interior
angle, "b" is 180-114=66 degrees.

The x component is 39*cos(66) = 15.862729080 and
the y component is 39*sin(66) = 35.628272848.

Since this vector is in quadrant 4, the
coordinates would be (15.86, -35.63).

x(final) = -270.82 + 15.86 = -254.96
y(final) = -359.39 + -35.63 = -395.02

Converting back to polar coordinates:

magnitude = sqrt(-254.96^2 + -395.02^2)
magnitude = sqrt(65004.6016 + 156040.8004)
magnitude = 470.15

angle = arctan(-395.02/-254.96)
angle = arctan(1.549341)
angle = 237.16 degrees

Therefore, the plane will end up flying at 470 mph
at a heading of 237.16 degrees

-Sorry I know its a lot!

anonymous · Answer

If someone could help me with the drawing that would help the most..