Bearings and scale drawingIB MYP Maths Standard: Revision notes

A bearing is a direction given as an angle measured clockwise from north. It is always written with three figures, so $72^\circ$ is written $072^\circ$. North is $000^\circ$, east $090^\circ$, south $180^\circ$ and west $270^\circ$. To find the bearing of B from A, stand at $A$, face north, then turn clockwise until you face $B$. The word after "from" tells you where to put the north line.

If the bearing of $B$ from $A$ is $\theta$, the bearing of $A$ from $B$ is the back bearing. The two north lines are parallel, so co-interior angles add to $180^\circ$.

• If $\theta<180^\circ$, back bearing $=\theta+180^\circ$.
• If $\theta\geq180^\circ$, back bearing $=\theta-180^\circ$. Example: $A$ to $B$ is $072^\circ$, so $B$ to $A$ is $072^\circ+180^\circ=252^\circ$. Pattern: the two bearings always differ by exactly $180^\circ$.

A scale such as $1:50\,000$ means $1$ unit on the map is $50\,000$ of the same unit in real life. So $1$ cm on the map is $50\,000$ cm $=500$ m $=0.5$ km.

• Map to real: multiply by the scale number, then convert the units.
• Real to map: convert to the same units, then divide by the scale number. Example: $6.4$ cm on a $1:50\,000$ map is $6.4\times50\,000=320\,000$ cm $=3.2$ km. If the scale changes to $1:25\,000$ the same distance looks twice as long: $12.8$ cm.

To make a scale drawing, choose a scale so that the drawing fits the page, for example $1$ cm to $2$ km. Draw the north line at each turning point, measure each bearing clockwise with a protractor and mark each distance with a ruler. Then measure the unknown distance or angle from your drawing and convert back using the scale. A drawing gives an answer that is only as accurate as your measuring, usually to the nearest millimetre or degree, so give the answer to a sensible accuracy.

Most bearing problems create a right-angled triangle, because north, east, south and west are at $90^\circ$ to each other. Sketch the journey, add north lines, and find the angle inside the triangle first. Example: a boat sails $12$ km east, then $5$ km north. Distance back $=\sqrt{12^2+5^2}=13$ km. The angle at the start is $\tan^{-1}\frac{5}{12}=22.6^\circ$ north of east. So the bearing of the finish is $090^\circ-22.6^\circ=067^\circ$ (nearest degree), and the back bearing is $67.4^\circ+180^\circ=247^\circ$. In real-life (criterion D) problems, state assumptions such as constant speed and straight paths, and say how they affect how accurate your answer is.