# Thread: vectors

1. ## vectors

An aircraft flies due east from A to B where AB=200 km . The wind is blowing from the direction 030 at 60 km/h . THe speed of the aircraft in still air is 300 km/h and the pilot sets the course on the bearing x . Find x

Can someone check my diagram and this table as well ,

course velocity = 300 and the direction is due east

wind velocity = 60 with a bearing of 30

track velocity = (speed , don know ) , and at a bearing of x .

i used the cosine rule and found that AB=334 km/h and x=81.1 degree .

However , the correct answer is 80 degree . Where did i go wrong ??

2. Hello hooke
Originally Posted by hooke
An aircraft flies due east from A to B where AB=200 km . The wind is blowing from the direction 030 at 60 km/h . THe speed of the aircraft in still air is 300 km/h and the pilot sets the course on the bearing x . Find x

Can someone check my diagram and this table as well ,

course velocity = 300 and the direction is due east

wind velocity = 60 with a bearing of 30

track velocity = (speed , don know ) , and at a bearing of x .

i used the cosine rule and found that AB=334 km/h and x=81.1 degree .

However , the correct answer is 80 degree . Where did i go wrong ??
A couple of mistakes:
The wind is blowing from $\displaystyle 030^o$. So you've got the arrow in the wrong direction.

In your diagram, you've marked the ground-speed as $\displaystyle 300$, rather than the speed of the aircraft through the air.
So the diagram should look like the one I've attached.

If you use the Sine Rule on this triangle, you'll get:
$\displaystyle \frac{\sin(90^o-x)}{60}=\frac{\sin120^o}{300}$
This will give $\displaystyle x = 80^o$.

Grandad

3. Originally Posted by Grandad
Hello hookeA couple of mistakes:
The wind is blowing from $\displaystyle 030^o$. So you've got the arrow in the wrong direction.

In your diagram, you've marked the ground-speed as $\displaystyle 300$, rather than the speed of the aircraft through the air.
So the diagram should look like the one I've attached.

If you use the Sine Rule on this triangle, you'll get:
$\displaystyle \frac{\sin(90^o-x)}{60}=\frac{\sin120^o}{300}$
This will give $\displaystyle x = 80^o$.

Grandad
thanks , one question about the direction of the wind , it only says from 030 , but it doesn't say from where .. ?? i am still not sure how to intepret this ? could u explain .

4. Originally Posted by hooke
thanks , one question about the direction of the wind , it only says from 030 , but it doesn't say from where .. ?? i am still not sure how to intepret this ? could u explain .
I'm not sure what you mean by "from where". Wind does not come from a particular point and vectors can be "moved" anywhere you want as long as you keep the direction and length the same. Saying that the wind is blowing "from 30 degrees" means that at every point the velocity vector is pointing toward 180+ 30= 210 degrees on the compass. Granddad drew it at the tip of the planes airspeed vector because because he was adding the vectors.

5. Originally Posted by HallsofIvy
I'm not sure what you mean by "from where". Wind does not come from a particular point and vectors can be "moved" anywhere you want as long as you keep the direction and length the same. Saying that the wind is blowing "from 30 degrees" means that at every point the velocity vector is pointing toward 180+ 30= 210 degrees on the compass. Granddad drew it at the tip of the planes airspeed vector because because he was adding the vectors.
A big thank you to both of you !!