Bearings and scale drawings
🎯 In this topic you will
- Use bearings as a measure of direction.
- Use bearings and scaling to interpret position on maps and plans.
🧠 Key Words
- bearing
Show Definitions
- bearing: A direction written as a three-figure clockwise angle measured from north (for example, 065°, 240°).
A bearing describes the direction of one object from another.
It is an angle measured from north in a clockwise direction.
A bearing can have any value from $0^\circ$ to $360^\circ$. It is always written with three figures.
In this diagram, the bearing from A to B is $120^\circ$.
In this diagram, the bearing from A to B is $065^\circ$.
❓ EXERCISES
1. For each diagram, write the bearing of B from A.
Use a protractor to measure the angle from north in a clockwise direction.
👀 Show answer
a.$065^\circ$ b.$140^\circ$ c.$225^\circ$ d.$300^\circ$.
2. Draw diagrams similar to those in Question $1$, to show these bearings of B from A.
a. $025^\circ$ b. $110^\circ$ c. $195^\circ$ d. $330^\circ$
👀 Show answer
a.$025^\circ$, b.$110^\circ$, c.$195^\circ$, d.$330^\circ$. Label B on each ray.
3. This is part of Freya’s homework.
Is Freya correct? Explain your answer.
👀 Show answer
❓ EXERCISES
🧠 Tip
To find the bearing of the shop from the school (part a) you need to measure the angle at the school. To find the bearing of the school from the shop (part b) you need to measure the angle at the shop.
4. The diagram shows the positions of a shop and a school.
a. Write the bearing of B (the shop) from A (the school).
b. Write the bearing of the school from the shop.
👀 Show answer
a. From the school to the shop: approximately $035^\circ$ (your value may vary slightly depending on measurement).
b. Reverse direction from the shop to the school: add $180^\circ$ → approximately $215^\circ$.
🧠 Think like a Mathematician
5. The diagram shows the position of a tree and a lake. Seren, Taylor and Ros are standing at the tree.
Seren walks straight from the tree to the lake.
a. On what bearing must she walk?
Taylor walks north from the tree. After a short distance she then walks to the lake.
b. Is the bearing she walks on to the lake, larger or smaller than the bearing from the tree to the lake? Explain your answer.
Ros walks south from the tree. After a short distance she then walks to the lake.
c. Is the bearing she walks on to the lake, larger or smaller than the bearing from the tree to the lake? Explain your answer.
d. What can you say about how bearings change as you move north or south from the original position before turning to walk towards another object?
👀 Show Answer
- a. Bearing of the lake from the tree is an acute angle measured clockwise from north (about $035^\circ$ from the sketch).
- b.Larger. After moving north, the north–south distance to the lake decreases while the east–west distance is similar, so the ratio $\frac{\text{east}}{\text{north}}$ increases. The bearing (clockwise from north) therefore increases, moving closer to $090^\circ$.
- c.Smaller. After moving south, the north–south distance to the lake increases, the ratio $\frac{\text{east}}{\text{north}}$ decreases, and the bearing decreases, moving closer to $000^\circ$.
- d. For a destination to the east of you, moving north increases the bearing (towards $090^\circ$), while moving south decreases it (towards $000^\circ$). If the destination were to the west, the trends would reverse.
❓ EXERCISES
6. Arun goes for a walk. The diagram shows Arun’s initial position (A), a farm (F), a pond (P), a tree (T) and a bridge (B). Write the bearing Arun follows to walk from
a. A to F
b. F to P
c. P to T
d. T to B
e. B to A
👀 Show answer
a.$050^\circ$ b.$165^\circ$ c.$260^\circ$ d.$335^\circ$ e.$120^\circ$.
🧠 Think like a Mathematician
7. Work independently to answer these questions.
a. For each diagram, write the bearing of Y from X and X from Y.
b. Draw two different diagrams of your own, plotting two points X and Y. In each diagram, the bearing of Y from X must be less than $180^\circ$. For each of your diagrams, write the bearing of Y from X and of X from Y.
c. What do you notice about each pair of answers in parts a and b?
d. Copy and complete this rule for two points X and Y, when the bearing of Y from X is less than $180^\circ$:
When the bearing of Y from X is $m^\circ$, the bearing of X from Y is ..............$^\circ$.
e. Reflect on your answers to parts c and d. What can you say about reverse bearings?
👀 Show Answer
- a (sample measurements)
i Y from X: about $030^\circ$; X from Y: about $210^\circ$.
ii Y from X: about $135^\circ$; X from Y: about $315^\circ$.
iii Y from X: about $060^\circ$; X from Y: about $240^\circ$.
(Your exact values may differ slightly depending on measurement.) - b Example pairs you could draw:
Case 1: Y from X $= 070^\circ$ ⇒ X from Y $= 250^\circ$.
Case 2: Y from X $= 150^\circ$ ⇒ X from Y $= 330^\circ$. - c The two bearings in each pair differ by $180^\circ$ (and add to $360^\circ$).
- d When Y from X is $m^\circ$ (with $m<180$), then X from Y is $m+180^\circ$.
- e Reverse bearings are always the original bearing plus (or minus) $180^\circ$, reduced to a three-figure value between $000^\circ$ and $359^\circ$.
❓ EXERCISES
This is part of Marcus’s homework.
Marcus uses alternate angles to work out the bearing of A from B.
8. For each diagram
i write the bearing of B from A
ii use Marcus’s method to work out the bearing of A from B.
👀 Show answer
a. i B from A: $077^\circ$. ii A from B: $077^\circ + 180^\circ = 257^\circ$.
b. i B from A: $118^\circ$. ii A from B: $118^\circ + 180^\circ = 298^\circ$.
c. The given angle is at B, from north to BA.
ii A from B: $016^\circ$. i B from A (reverse): $016^\circ + 180^\circ = 196^\circ$.
❓ EXERCISES
9. This is part of Sofia’s homework.
Sofia uses alternate angles to work out the bearing of Q from P. For each diagram
i write the bearing of P from Q
ii use Sofia’s method to work out the bearing of Q from P.
👀 Show answer
a.i P from Q: $244^\circ$. ii Q from P: $244^\circ - 180^\circ = 064^\circ$.
b.i P from Q: $348^\circ$. ii Q from P: $348^\circ - 180^\circ = 168^\circ$.
c.i P from Q: $204^\circ$. ii Q from P: $204^\circ - 180^\circ = 024^\circ$.
You know that you can write a scale in three ways:
- using the word ‘represents’, for example, ‘$1$ cm represents $100$ cm’
- using the word ‘to’, for example, ‘$1$ to $100$’
- using a ratio sign, for example, ‘$1\!:\!100$’
Also you know that:
- a bearing describes the direction of one object from another
- a bearing is an angle measured from north in a clockwise direction
- a bearing can have any value from $0^\circ$ to $360^\circ$
- a bearing is always written with three figures.
You can use bearings in scale drawings to help you to solve problems. When you make a scale drawing, make sure you always measure the lengths and angles accurately.
❓ EXERCISES
1. Firash walks on a bearing of $050^\circ$ for $800$ m.
Copy and complete the working and scale drawing to show Firash’s journey.
Use a scale of $1$ cm represents $100$ m.
Distance on scale drawing = $800 \div 100$ = [ ] cm
👀 Show answer
Draw a north arrow at the start, measure a clockwise bearing of $050^\circ$, then draw a straight line of length $8$ cm on that bearing.
2. Jahia and Rafiki stand by a tree.
Jahia walks on a bearing of $140^\circ$ for $50$ m.
Rafiki walks on a bearing of $230^\circ$ for $70$ m.
a. Show both their journeys on the same scale drawing.
Use a scale of $1$ cm represents $10$ m.
b. On your scale drawing, measure the distance between the girls at the end of their walk.
How far apart are the girls in real life?
👀 Show answer
b. The angle between the two bearings is $230^\circ - 140^\circ = 90^\circ$, so the endpoints form a right triangle with legs $50$ m and $70$ m.
Distance between the girls: $\sqrt{50^2 + 70^2} = \sqrt{7400} \approx 86$ m (about $8.6$ cm on the scale drawing).
🧠 Think like a Mathematician
3. Work independently to answer this question.
- A yacht is $70$ km west of a speedboat.
- The yacht sails on a bearing of $082^\circ$.
- The speedboat travels on a bearing of $285^\circ$.
- Could the yacht and the speedboat meet? Explain your answer.
Method (coordinate or scale drawing):
- Place the speedboat at the origin and the yacht at $(-70,0)$ (east–north axes; km).
- Bearings convert to direction vectors using $(\sin\theta,\ \cos\theta)$. Draw rays or write parametric lines for each vessel.
- Either measure on a scale diagram to see if the rays cross, or solve the two linear equations to find an intersection point.
- If the paths cross, compare distances along each path to comment on whether they can arrive at the same time for typical speeds.
👀 Show Answer
Solve $(-70,0)+t(\sin82^\circ,\ \cos82^\circ) = u(\sin285^\circ,\ \cos285^\circ)$ with $t,u\ge0$. This gives $t\approx46.36$ km and $u\approx24.93$ km, meeting at about $(-24.1,\ 6.45)$ km (i.e., 24.1 km west and 6.45 km north of the speedboat’s start).
Therefore the paths do intersect, so they could meet if their timings/speeds allow. If they start together, they meet provided the speed ratio satisfies $\dfrac{v_{\text{speedboat}}}{v_{\text{yacht}}}=\dfrac{u}{t}\approx0.537$ (e.g., the speedboat travels at about $54\%$ of the yacht’s speed, or one starts later). If they both travel at the same speed and start simultaneously, they arrive at different times, so they would not meet.
❓ EXERCISES
🧠 Tip
Remember: ‘$1:2\,000\,000$’ means $1$ cm on the diagram represents $2\,000\,000$ cm on the ground, so start by changing $2\,000\,000$ cm into km.
4. Two lighthouses are $160$ km apart.
Lighthouse A is north of lighthouse B.
A ship is on a bearing of $152^\circ$ from lighthouse A and $042^\circ$ from lighthouse B.
Draw a scale diagram to show the position of the ship.
Use a scale of $1:2\,000\,000$.
👀 Show answer
Scale:$2\,000\,000$ cm $=$$20$ km, so $1$ cm on the diagram represents $20$ km.
Step 1. Mark lighthouse B and draw a vertical line upwards to lighthouse A with length $160 \div 20 = 8$ cm (A north of B).
Step 2. From A draw a ray at bearing $152^\circ$ (clockwise from north). From B draw a ray at bearing $042^\circ$.
Step 3. Their intersection is the ship’s position. Label the point “Ship”. (You may measure its distances from A or B if required.)
❓ EXERCISES
5. This is part of Teshi’s classwork.
a. Is Teshi’s sketch correct? Explain your answer.
b. Show that Yue must walk $8.2$ km on a bearing $137^\circ$.
👀 Show answer
a.No. From Yue to the lake the bearing must be between $090^\circ$ and $180^\circ$ (towards the south-east), so the path from Yue should slope downwards from Yue. Teshi’s dashed line is drawn towards the north-east, which does not match the required bearing.
b. Place Yue at the origin $(0,0)$ and Jun at $(0,-8)$ (kilometres). Jun walks $6$ km on a bearing $070^\circ$. Components from Jun to the lake are east $=6\sin70^\circ\approx6(0.9397)=5.638$ and north $=6\cos70^\circ\approx6(0.3420)=2.052$. So the lake is at $(5.638,\ -8+2.052)=(5.638,\ -5.948)$.
Distance from Yue to the lake: $d=\sqrt{(5.638)^2+(-5.948)^2}\approx\sqrt{31.78+35.38}\approx\sqrt{67.16}\approx8.2$ km.
Bearing of the lake from Yue (clockwise from north) is $\theta=\operatorname{atan2}(E,N)=\operatorname{atan2}(5.638,\,-5.948)\approx136.9^\circ$, which to the nearest degree is $137^\circ$.
🧠 Think like a Mathematician
6. Work independently to answer this question.
A ship leaves a harbour and sails $80$ km on a bearing of $120^\circ$. The ship then sails $100$ km on a bearing of $030^\circ$.
a. Make a scale drawing of the ship’s journey. Choose a sensible scale.
b. What distance must the ship now sail to return to the harbour?
c. On what bearing must the ship now sail to return to the harbour?
d. Reflect: compare methods (scale drawing vs. calculation) and note any common mistakes and how to avoid them.
👀 Show Answer
- a (one good choice): Use scale $1$ cm $=$$10$ km. From the harbour draw a $120^\circ$ ray of length $8$ cm, then from its end a $030^\circ$ ray of length $10$ cm. Measure the straight-line distance and bearing from the final point back to the harbour.
- b (calculation): Resolve each leg into east–north components using bearings → direction $(\sin\theta,\ \cos\theta)$.
After first leg: $80(\sin120^\circ,\ \cos120^\circ)=(69.28,\,-40.00)$ km.
After second leg: add $100(\sin30^\circ,\ \cos30^\circ)=(50.00,\ 86.60)$ km.
Final displacement from harbour: $(E,N)=(119.28,\ 46.60)$ km.
Distance back: $\sqrt{119.28^2+46.60^2}\approx128$ km. - c: Bearing from harbour to the ship is $\operatorname{atan2}(E,N)\approx68.7^\circ$. Reverse bearing (ship → harbour) is $68.7^\circ+180^\circ\approx248.7^\circ$ → $249^\circ$ (three-figure).
- d (notes): Common slips: mixing up sine/cosine with east/north; forgetting bearings are measured clockwise from north; writing two-figure angles (e.g., $68^\circ$ instead of $068^\circ$); not redrawing a north arrow at the turning point on a scale diagram.
❓ EXERCISES
7. Mark leaves his tent and walks $12$ km on a bearing of $045^\circ$. He then walks $16$ km on a bearing of $275^\circ$.
a. Make a scale drawing of Mark’s walk. Use a scale of $1$ cm represents $2$ km.
b. How far must Mark now walk in a straight line to return to his tent?
c. On what bearing must Mark now walk to return in a straight line to his tent?
👀 Show answer
a. Scale: $1$ cm $=$$2$ km.
First leg: draw a $6$ cm line on bearing $045^\circ$ from the tent. Second leg: from its end, draw an $8$ cm line on bearing $275^\circ$ (clockwise from north).
b. Resolve components (east, north). First leg: $E_1=12\sin45^\circ\approx8.49$ km, $N_1=12\cos45^\circ\approx8.49$ km. Second leg: $E_2=16\sin275^\circ\approx-15.94$ km, $N_2=16\cos275^\circ\approx1.39$ km. Total from tent: $E=E_1+E_2\approx-7.45$ km (west), $N=N_1+N_2\approx9.88$ km (north). Distance back to tent: $d=\sqrt{E^2+N^2}\approx\sqrt{(-7.45)^2+(9.88)^2}\approx12.4$ km (which is about $6.2$ cm on the scale drawing).
c. Bearing from current position to the tent (clockwise from north): use $\theta=\operatorname{atan2}(E_{\text{back}},N_{\text{back}})$ with $E_{\text{back}}=+7.45$, $N_{\text{back}}=-9.88$. $\theta\approx142.97^\circ$ so the three-figure bearing is $143^\circ$.
❓ EXERCISES
🧠 Tip
In parts $b\,ii$, $c\,ii$ and $d\,ii$ make sure you give the real life distances.
8. The diagram shows two radio masts $P$ and $Q$.
a. Make a scale drawing of the diagram. Use a scale of $1\!:\!20\,000$.
b. A farmhouse is $8$ km on a bearing of $330^\circ$ from $P$.
i Draw the position of the farmhouse on your diagram.
ii Measure the distance and bearing of the farmhouse from $Q$.
c. A shop is $12$ km on a bearing of $210^\circ$ from $Q$.
i Draw the position of the shop on your diagram.
ii Measure the distance and bearing of the shop from $P$.
d. A cafe is on a bearing of $070^\circ$ from $P$ and $135^\circ$ from $Q$.
i Draw the position of the cafe on your diagram.
ii Measure the distance of the cafe from $P$ and $Q$.
👀 Show answer
a. Draw $P$. From $P$ mark a point $10$ km to the east; from there mark a point $5$ km to the north — this is $Q$. (Use the given scale.)
b.i From $P$, mark the farmhouse $8$ km on bearing $330^\circ$.
ii From $Q$ to the farmhouse: distance $\approx 14.13$ km; bearing $\approx 278^\circ$ (three figures).
c.i From $Q$, mark the shop $12$ km on bearing $210^\circ$.
ii From $P$ to the shop: distance $\approx 6.71$ km; bearing $\approx 143^\circ$.
d. The cafe is at the intersection of the rays: bearing $070^\circ$ from $P$ and $135^\circ$ from $Q$.
ii Distances: from $P$$\approx 11.70$ km; from $Q$$\approx 1.41$ km.
Note: Your measured values from a scale drawing should be very close to these; slight variation is normal.
9. A lighthouse is $75$ km east of a port.
The captain of a ship knows he is on a bearing of $315^\circ$ from the lighthouse.
He also knows he is on a bearing of $052^\circ$ from the port.
a. Draw a scale diagram to show the position of the ship. Use a scale of $1\!:\!1\,000\,000$.
b. How far is the ship from the lighthouse?
c. How far is the ship from the port?
👀 Show answer
a. Place the port and then the lighthouse $75$ km to the east. From the lighthouse draw a ray on bearing $315^\circ$; from the port draw a ray on bearing $052^\circ$. Their intersection is the ship.
b. Ship–lighthouse distance $\approx 46.5$ km.
c. Ship–port distance $\approx 53.4$ km.
❓ EXERCISES
10. Alicia is participating in a charity sailing race on the Mar Menor. The map shows the route she takes.
a. On what bearing must she sail to get from
i the start to A ii A to B iii B to the finish.
The map has a scale of $1:250\,000$.
b. What is the total distance that Alicia sails?
Alicia earns $\$56$ for charity, for every kilometre she sails.
c. What is the total amount that Alicia raises for charity?
👀 Show answer (a)
i. start → A: about $110^\circ$
ii. A → B: about $240^\circ$
iii. B → finish: about $350^\circ$.
👀 Show answer (b)
Measure each leg on the map (in cm): let the lengths be $\ell_1$ (start→A), $\ell_2$ (A→B), $\ell_3$ (B→finish).
Convert to real distances: $d_i=2.5\,\ell_i$ km, and total distance $D = 2.5(\ell_1+\ell_2+\ell_3)$ km.
Example (for guidance): if $\ell_1=4.0$ cm, $\ell_2=4.8$ cm, $\ell_3=3.6$ cm, then $D=2.5(4.0+4.8+3.6)=2.5\times12.4\approx31.0$ km.
👀 Show answer (c)
Using the example above with $D\approx31.0$ km, $\$56\times31.0\approx\$1\,736$ (to the nearest dollar). Replace with your measured total distance.