Having looked at true bearing and compass bearings, let us now look at some problems in bearings

Problem 1: Two people start sailing from the same point. The first person sails due west for 40 km and the second person sails round 230° until he is due south of the first person. How far did the second person sail?

The sailing of the two persons (let’s call them A and B respectively) is shown in the diagram here.

Bearing of B from 0 is 230°

270 – 230 = 40°, or W40°S

 

Looking specifically at the right angled triangle OAB,

x = hypotenuse, and adjacent side = 40 km

cos 40° =

x = 

=  52.22 km (or 52 km to the nearest km).

 

Problem 2: Town A is 4 km north and 3 km east of town B. Find the bearing of town A from town B.

tan θ = = 

θ =  

=  36.8698 = 36° 52′

So bearing of A from B will be 036° 52′ or N36° 52’E

 

Problem 3: Two towns A and B are 10 km apart, and the bearing of B from A is 300°. Find how far B is west of A.

The above scenario can be represented by the diagram shown on the left below. On the right, we look specifically at the right angled triangle ABD.

sin 60° =

sin 60° =

x = 10 x sin 60°

=  8.66 km

So town B is 8.66 km west of town A