An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. how many degrees off course will the plane end up flying, and what is the plane's speed relative to the ground?
So here is my work...
50cos(45)+500= 535.355
50cos(45)= 35.355
...
sqrt(535.355^2+35.355^2)= 536.521
arctan(35.355/535.355)= 3.778 degrees
...
So my answer would be 536.521 km/hr and 3.778 degrees right?
2 answers
The wind is blowing to the SE. That will make the resultant speed less than 500 km/hr. That will make your final course a little more eastward than your calculation.
As Steve pointed out, you answer makes no sense, nor can I understand what your calculations are supposed to represent.
It almost looks like you are assuming there is a right-angled triangle. There isn't one.
If you make a sketch for the problem you would see that this is a clear-cut case of the cosine law.
(ground speed)^2 = 500^2 + 50^2 - 2(50)(500)cos45
...
ground speed = 465.99 km/h or 466 km/h
for the angle, use the sine law :
sinØ/50 = sin45/465.99
sinØ = .07587...
Ø = 4.35°
It almost looks like you are assuming there is a right-angled triangle. There isn't one.
If you make a sketch for the problem you would see that this is a clear-cut case of the cosine law.
(ground speed)^2 = 500^2 + 50^2 - 2(50)(500)cos45
...
ground speed = 465.99 km/h or 466 km/h
for the angle, use the sine law :
sinØ/50 = sin45/465.99
sinØ = .07587...
Ø = 4.35°
2 answers