- Line the zero of the ruler up with the start of the line — not the end of the plastic.
- Read to the nearest millimetre and write the answer in the units asked for.
- For an angle, put the protractor's centre on the vertex and the zero line along one arm.
- Read from the scale that starts at zero on that arm.
- Sense-check: an acute-looking angle must read less than $90^\circ$.
| You know… | You want… | Do this |
|---|---|---|
| Map / drawing length | Real length | Multiply by the scale number |
| Real length | Map / drawing length | Divide by the scale number |
On a $1 : 50\,000$ map, two villages are $9.2$ cm apart. Find the real distance in kilometres.
A room is $7.5$ m long. How long is it on a plan drawn to a scale of $1 : 50$?
- Identify the point you are measuring from — it comes after the word "from".
- Draw a North line at that point.
- Measure clockwise from North to the line joining the two points.
- Write the answer with three figures and a degree sign.
| Direction | Bearing | Direction | Bearing |
|---|---|---|---|
| North | $000^\circ$ | South | $180^\circ$ |
| North-East | $045^\circ$ | South-West | $225^\circ$ |
| East | $090^\circ$ | West | $270^\circ$ |
| South-East | $135^\circ$ | North-West | $315^\circ$ |
If the bearing is more than $180^\circ$, subtract $180^\circ$
(a) The bearing of $Q$ from $P$ is $073^\circ$. Find the bearing of $P$ from $Q$.
(b) The bearing of $S$ from $R$ is $296^\circ$. Find the bearing of $R$ from $S$.
- Choose a scale that makes the drawing fit the page.
- Convert each real distance into a drawing length by dividing.
- Mark the starting point and draw a North line there.
- Measure the first bearing clockwise from North and draw the leg to the correct length.
- Draw a fresh North line at the new point before measuring the next bearing.
- Measure the final answer from the drawing and multiply back up to real life.
A boat sails $40$ km on a bearing of $060^\circ$, then $30$ km on a bearing of $150^\circ$. Using a scale of $1$ cm : $10$ km, find the distance and bearing of the boat from its start.
Measuring the drawing should give $5$ cm and about $097^\circ$ — matching the calculation ✓
The bearing of $B$ from $A$ is $130^\circ$. The bearing of $C$ from $A$ is $210^\circ$. Find the angle $BAC$.
Bearings
From North, clockwise, three figures.
"B from A"
Stand at $A$; the North line goes at $A$.
Back bearing
Add $180^\circ$ if under $180^\circ$; subtract if over.
Compass points
N $000$, NE $045$, E $090$, SE $135$, S $180$, SW $225$, W $270$, NW $315$.
Map → real
Multiply by the scale, then convert cm → m → km.
Real → map
Convert to cm first, then divide by the scale.
New North lines
Draw a fresh one at every turning point.
Angle between bearings
From the same point, just subtract.
Tolerance
Usually $\pm 2$ mm and $\pm 2^\circ$.
Write these compass directions as three-figure bearings: (a) South-East, (b) West, (c) North.
▶ Show solution
(a) $135^\circ$ (b) $270^\circ$ (c) $000^\circ$ (or $360^\circ$)
The bearing of $Y$ from $X$ is $048^\circ$. Find the bearing of $X$ from $Y$.
▶ Show solution
$048^\circ$ is less than $180^\circ$, so add $180^\circ$.
$48 + 180 = 228^\circ$
The bearing of $M$ from $L$ is $315^\circ$. Find the bearing of $L$ from $M$.
▶ Show solution
$315^\circ$ is more than $180^\circ$, so subtract $180^\circ$.
$315 - 180 = 135^\circ$
On a map with scale $1 : 20\,000$, a river is $12.5$ cm long. Find its real length in kilometres.
▶ Show solution
$12.5 \times 20\,000 = 250\,000$ cm
$250\,000 \div 100 = 2500$ m $= 2.5$ km
A garden is $24$ m long. It is drawn to a scale of $1 : 200$. How long is it on the plan?
▶ Show solution
$24$ m $= 2400$ cm
$2400 \div 200 = 12$ cm
The bearing of $B$ from $A$ is $035^\circ$ and the bearing of $C$ from $A$ is $128^\circ$. Find angle $BAC$.
▶ Show solution
Both are measured from the same North line at $A$, so subtract:
$\angle BAC = 128 - 35 = 93^\circ$
A walker heads $5$ km due North, then $12$ km due East. Find (a) the direct distance from the start, (b) the bearing of the finish from the start, to the nearest degree.
▶ Show solution
(a) The two legs are perpendicular, so use Pythagoras:
$d = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ km
(b) $\tan\theta = \dfrac{12}{5} = 2.4$, so $\theta = 67.4^\circ$ measured from North.
Bearing $= \mathbf{067^\circ}$ (nearest degree).
A ship sails $18$ km on a bearing of $090^\circ$, then $18$ km on a bearing of $180^\circ$. Find its distance and bearing from the start.
▶ Show solution
$090^\circ$ is due East and $180^\circ$ is due South, so the legs are perpendicular.
Distance $= \sqrt{18^2 + 18^2} = \sqrt{648} = 25.5$ km (1 d.p.)
Equal East and South components means the direction is exactly South-East.
Bearing $= \mathbf{135^\circ}$
A walker uses a $1 : 25\,000$ map. The route measures $22.4$ cm. She walks at $5$ km/h. How long will the walk take, to the nearest minute?
▶ Show solution
Real distance $= 22.4 \times 25\,000 = 560\,000$ cm $= 5600$ m $= 5.6$ km.
Time $= \dfrac{5.6}{5} = 1.12$ hours.
$0.12 \times 60 = 7.2$ minutes.
About $1$ hour $7$ minutes.
Town $B$ is $60$ km from town $A$ on a bearing of $120^\circ$. Town $C$ is $60$ km from $B$ on a bearing of $240^\circ$.
(a) Find angle $ABC$. (b) What type of triangle is $ABC$? (c) Find the distance $AC$. (d) Find the bearing of $C$ from $A$.
▶ Show solution
(a) The bearing of $A$ from $B$ is the back bearing of $120^\circ$: $120 + 180 = 300^\circ$.
At $B$, the two directions are $300^\circ$ (towards $A$) and $240^\circ$ (towards $C$).
$\angle ABC = 300 - 240 = 60^\circ$
(b) $AB = BC = 60$ km, so the triangle is isosceles. With the apex angle $60^\circ$, the base angles are $(180-60)\div2 = 60^\circ$ each — so it is equilateral.
(c) All sides equal, so $AC = \mathbf{60}$ km.
(d) At $A$, the direction to $B$ is $120^\circ$ and $\angle BAC = 60^\circ$.
$C$ lies to the west of the line $AB$ (since the second leg turned back), so the bearing of $C$ from $A$ is $120 - 60 = \mathbf{060^\circ}$.