Back Bearings

Worked example

1. 1

   $064^{\circ }<180^{\circ }$ — Check the size first. 064° is less than half a turn, so adding 180° will still give less than 360°.

2. 2

   $064^{\circ }+180^{\circ }=244^{\circ }$ — Half a turn, 180°, points the line exactly the other way. The bearing of A from B is 244°.

Worked example

1. 1

   $290^{\circ }\ge 180^{\circ }$ — 290° is more than half a turn, so adding 180° would go past 360°.

2. 2

   $290^{\circ }-180^{\circ }=110^{\circ }$ — Take away 180° instead. It is still half a turn, just the other way round. The bearing of A from B is 110°.

Worked example

1. 1

   $\text{at }A:64^{\circ }\text{ from north}$ — The bearing of B from A is 064°, so the angle between A's north line and AB is 64°.

2. 2

   $\text{north at }A\parallel \text{north at }B$ — North is the same direction everywhere, so the two north lines are parallel. That lets you use angle facts.

3. 3

   $\text{at }B:180^{\circ }-64^{\circ }=116^{\circ }$ — The angle between B's north line and BA sits on the same side of AB as the 64°. They are co-interior angles, which add to 180°.

4. 4

   $360^{\circ }-116^{\circ }=244^{\circ }$ — The bearing at B turns clockwise from north to BA, which is the outside of the 116° angle. Angles at a point add to 360°.

Worked example

1. 1

   $070^{\circ }+180^{\circ }=250^{\circ }$ — First the bearing of A from B. 070° is under 180°, so add 180°. This gives the direction of BA, measured at B.

2. 2

   $\text{angle }ABC=80^{\circ }$ — The angle inside the triangle at B, between BA and BC. C is in the south-east quarter from B, so turning from BA to BC goes anticlockwise, from 250° down towards 180°.

3. 3

   $250^{\circ }-80^{\circ }=170^{\circ }$ — Turning anticlockwise makes a bearing smaller, so take the 80° away.

4. 4

   $\text{bearing of }C\text{ from }B=170^{\circ }$ — Check: 170° is between east (090°) and south (180°), so C is in the south-east quarter, as the question says.