Week 11

Interpret and use bearings

The emphasis in these task is not so much on calculating specific bearings than on visualising them, especially in situations where the bearings change.

MONDAY: Here we have two lighthouses roughly north of Rod' s boat. However, as their bearings change, with one increasing and one decreasing, one of them must be somewhat to the east of north and the other somewhat west of north. Which is which?

Can students visualise the situation? If not, they might find it helpful to make a sketch. As we can see (below), as Rod sails northwards the red angle increases and the green angle decreases, and so A is somewhat to the right (east) of Rod, and B is somewhat to the left (west).

 

TUESDAY: Here we have a lighthouse which is initially south west of a boat. A minute later it lies slightly more to the west of the boat. So where might the boat be at that point in time?

Again, it is worth trying to visualise what is going on, but at some stage a sketch might be useful, if only to verify students' claims. The task can be solved quite effectively using a trial and improvement approach: it should soon become apparent that if the boat is at point B (below) when the bearing of the lighthouse is 135˚, the bearing will increase if the boat sails to any point to the east and/or south of the blue line (ie on any course east of north east and east of south west.

 

 

WEDNESDAY: Here we have a ship which at 12:05 is north and slightly east of a lighthouse. The ship travels in a north easterly direction and so will lie further and further to the east of the lighthouse and so the bearing will increase - but at what kind of rate?

Rather than focussing on what happens to the ship in the next few minutes, it is worth considering what happens as the ship keeps sailing north east (but assuming the surface of the sea lies on a plane, not on a sphere!). The bearing of the ship from the lighthouse will get closer and closer to 045˚, but never reach it. This means that for a given interval of time (5 minutes, say) the bearing increases in successively smaller amounts - the relation is not linear.

So if the bearing increases by 5˚ in the 5 minutes from 12:05 to 12:10, it will increase by less than 5˚ in the next 5 minutes. So at 12:15 the bearing will be (slightly) less that 025˚, and at 13:00 it will be (substantially?) less that 015˚ + 11×5˚ = 70˚, indeed, as argued above, it will be (somewhat) less than 045˚.

Again, a diagram might help (below). Here it turns out (by measuring the angles in the diagram) that the bearing is about 023˚ rather than 025˚ at 12:15, and about 036˚ at 13:00.

THURSDAY: This is a relatively open task, in as much as there are infinitely many solutions. However, the relation between the bearings is fairly simple.

There is a nice 'symmetrical' relation between the desired bearings (not surprisingly, perhaps). This is one way of expressing it: the sum of the two bearings is 600˚. We can see the relation from this diagram (below). So if a = 10, say, the pair of bearings would be 290˚ and 310˚.

FRIDAY: This task is perhaps a bit more down to earth. It asks for the value of a specific bearing.

How obvious is it that the bearing is equal to 205˚ + 90˚? One way to see this is to imagine a second bowling green, as in this diagram: