An aeroplane flies 120km on a bearing of 130 degrees. It then alters course and flies 250km on a bearing of 035 degrees.
a) How far east is the plane from its starting point?
b) How far north is the plane from its starting point?
c) How far is the plane from its starting point?
:confused:

Draw it out. You need to use trig (sin, because, tan) to find the other side.

Draw a picture... always. Did you?

I will show you how to do this so you can do more like it. One must be able to visualize the triangles formed.

The bearings, technically, are azimuths.

An azimuth is turned clockwise from north (y-axis) right around the circle.

The first is 120 km at 130 degrees.

So, the coordinates of the first point where he turns is

x=120sin(130) and y=120cos(130)

Now, continue from there and add what you get to these points to find the coordinates of the next point he goes to.

x=120sin(130)+250sin(35) and y=120cos(130)+250cos(35)

Now, you can use Pythagoras and these coordinates to answer part c.

See what I done?

p,q,r are three points on level ground bearing of r from p is 135 pqr is 55 and prq is 48 .calculate
1 p from r.
2 q from r
3 p from q.

Why did you post in an old one from months back? Please start your own thread. Folks are more likely to see it.

This one is too hard. Soz but not sure