1. Measured from North — North is always the $000°$ reference line.
2. Measured clockwise — never anticlockwise.
3. Three-figure notation — always written with exactly three digits, so $5°$ becomes $005°$ and $59°$ becomes $059°$.

Because the North lines at two different points are parallel, the bearing of B from A and the bearing of A from B are always exactly $180°$ apart:

Example: the bearing of A from B is $310°$. Since $310\ge180$, the reverse (B from A) is $310-180=130°$.

The two parallel North lines mean the connecting path AB acts as a transversal. The angle at one end and the equivalent angle at the other are linked by alternate and co-interior angle facts — specifically, co-interior angles sum to $180°$, which is exactly where the $\pm180°$ shortcut comes from. If a question doesn't fit the simple rule neatly (e.g. it mixes quadrants), fall back on explicitly finding the interior angle via co-interior or alternate angles, then build up the full bearing from there using $90°$ or $180°$ segments.

1. Convert the real distance to a drawing distance using the scale (e.g. $1\text{ cm}:20\text{ m}$ means $140\text{ m}\div20=7\text{ cm}$).
2. Draw a North line at the start point.
3. Centre your protractor on the start point, align $0°$ with North, and measure the bearing clockwise.
4. Measure the drawing distance along that line with a ruler and mark the point clearly with a cross.

Draw a new North line at whichever point you're measuring from — don't reuse the North line drawn at a different point. Measure the angle clockwise from that new North line, and convert any measured drawing distance back to a real distance by multiplying by the scale factor.

Draw a fresh North line at every turning point. To find the angle at B between the paths BA and BC, you typically need the reverse bearing of A from B first (using the rule above), then combine it with the given bearing of C from B — often via subtraction — to get the interior angle of the triangle ABC at that vertex. From there, the angle sum of a triangle ($180°$) can finish the problem.

The bearing of a lighthouse L from a harbour H is $148°$. A boat sails from H on this bearing for 6 km to reach L, then turns and sails back to H. What is the bearing of H from L?

1. The known bearing is $148°$, which is less than $180°$.
2. Reverse bearing $= 148+180=328°$.
3. The bearing of H from L is $328°$.