Question

7. An aircraft flies 400 km from a point O on a bearing of 025°

and then 700 km on a bearing of 080° to arrive at B.
a) How far north of O is B?
b) How far east of O is B?

Answer 1

• 484 km north
• 858 km east

Explanation:

You want the distances north and east of the origin that a plane ends up after flying 400 km at a bearing of 25°, then 700 km at 80°.

Coordinate transformation

For some distance r and bearing angle θ, the (north, east) coordinates will be ...

(north, east) = (r·cos(θ), r·sin(θ))

Application

For the trip at hand, the final coordinates of the plane are ...

a)

Distance north = (400 km)cos(25°) +(700 km)cos(80°)

= 362.52 km +121.55 km = 484.07 km

B is about 484 km north of O.

b)

Distance east = (400 km)sin(25°) +(700 km)sin(80°)

= 169.05 km +689.37 km = 858.41 km

B is about 858 km east of O.

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Additional comment

Rectangular coordinates of a point at distance r from the origin and an angle θ measured CCW from the +x axis are given by ...

(x, y) = (r·cos(θ), r·sin(θ))

You may notice the similarity to the coordinates described above. That is why we can use a calculator the way we have in the attachment. The imaginary part of the complex number represents the distance east.