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 $030^o$. So you've got the arrow in the wrong direction.

In your diagram, you've marked the ground-speed as $300$, 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:
$\frac{\sin(90^o-x)}{60}=\frac{\sin120^o}{300}$
This will give $x = 80^o$.

Hello hookeA couple of mistakes:
The wind is blowing from $030^o$. So you've got the arrow in the wrong direction.

In your diagram, you've marked the ground-speed as $300$, 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:
$\frac{\sin(90^o-x)}{60}=\frac{\sin120^o}{300}$
This will give $x = 80^o$.