G15: Measuring & Bearings
Measure line segments and angles; use bearings to specify direction
📋 Key Concepts
Measuring involves using instruments like rulers and protractors to find lengths and angles accurately.
Bearings are a way of describing direction using angles measured clockwise from North.
📝 Measuring Length
Using a ruler:
- Start measuring from zero mark, not the end of the ruler
- Read to the nearest millimetre
- Keep the ruler parallel to the line being measured
Example 1
Measure a line segment to the nearest mm. The ruler shows the line starts at 0 and ends between 4.5 cm and 4.6 cm.
Solution: Record as 4.5 cm or 45 mm (to nearest mm) or 4.55 cm (to nearest 0.5 mm)
📝 Measuring Angles
Using a protractor:
- Place the centre of the protractor on the vertex (corner point)
- Line up the base line with one of the angle's arms
- Read from zero, counting up to where the other arm crosses the scale
- Use the correct scale (inner or outer depending on which arm starts at zero)
Example 2
Measure an angle using a protractor. The angle opens to the right.
Solution:
Use the inner scale (starting from the right at 0°)
Read where the other arm crosses the scale
Tip: Estimate first - acute < 90°, obtuse > 90°
📝 Bearings
Three rules for bearings:
- Measure from North (the vertical line pointing up)
- Measure clockwise
- Write as a three-figure number (e.g., 035° not 35°)
Bearings are always measured clockwise from North
Example 3
What is the bearing of East from North?
Solution:
East is 90° clockwise from North.
Bearing = 090°
Example 4
What is the bearing of South-West?
Solution:
South is 180°, West is 270°
South-West is halfway: (180° + 270°) ÷ 2 = 225°
Bearing = 225°
📝 Finding Bearings
To find the bearing of B from A:
- Draw a North line at A
- Draw a line from A to B
- Measure the angle clockwise from North to the line AB
Example 5
Point B is directly South-East of point A. Find the bearing of B from A.
Solution:
South-East is 135° clockwise from North.
Bearing = 135°
📝 Back Bearings
Back bearing: The bearing to return from B to A. Add or subtract 180° from the original bearing.
Back bearing = Bearing ± 180°
Example 6
The bearing of B from A is 045°. Find the bearing of A from B.
Solution:
Back bearing = 45° + 180° = 225°
Or: 360° - 45° + 180° = 225°
Example 7
The bearing of Q from P is 280°. Find the bearing of P from Q.
Solution:
Back bearing = 280° - 180° = 100°
(Subtract 180° because the result would exceed 360° if we added)
🧮 Worked Examples — Exam Style
Bearings: clockwise from north, three figures. Distance questions combine bearings with trigonometry — draw the north line at BOTH points.
A walker walks 4 km north, then 3 km east. Find the bearing of the finish from the start (1 d.p.).
Triangle: north 4, east 3 → angle from north = tan⁻¹(3/4) = 36.9°
Bearing = 036.9° (three figures)
M1 for the tangent, A1 for the angle, B1 for three-figure form. The north-point line at the START is the reference.
Exam wording to expect: "measure/write the bearing of B from A", "find how far … has travelled north/east".
❓ Practice Questions
Q1: What is the bearing of North?
Q2: What is the bearing of West?
Q3: The bearing of B from A is 120°. What is the bearing of A from B?
Q4: What compass direction is bearing 315°?
Q5: A ship sails on a bearing of 070°. What bearing must it take to return?
Q6: A plane flies 200 km on a bearing of 130°. How far east has it travelled (3 s.f.)?
Q7: The bearing of B from A is 310°. Find the bearing of A from B.
✅ Answers
- 000°
- 270°
- 300° (120° + 180°)
- North-West
- 250° (070° + 180°)
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. 000° (north is the zero direction)
Q2. 270°
Q3. 120 + 180 = 300°
Q4. 315° is 45° west of north → north-west (NW)
Q5. 070 + 180 = 250°
Q6. East = 200 × sin 50° (130° is 50° past east of south... precisely: component east of north = 200 sin 130° = 200 sin 50°) = 153 km (3 s.f.)
Q7. 310 − 180 = 130°
📌 Key Facts to Memorise
- Bearings: measured CLOCKWISE from NORTH, always THREE digits (047°, not 47°)
- Back bearing: ±180° (add if < 180, subtract if ≥ 180)
- Compass: N 000°, NE 045°, E 090°, SE 135°, S 180°, SW 225°, W 270°, NW 315°
- North-line at BOTH points for two-way bearing problems
- Components: north = d cos θ, east = d sin θ (θ from north)
🗺️ In the Exam
Bearings: 2–5 marks on Papers 2/3 (protractor + ruler for measuring, calculator for trig versions). The three-figure convention is itself examined.
Links to: Pythagoras/trig (G20), scale drawings (R2), sine/cosine rules (G22).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always give bearings as three figures (e.g., 005°, not 5°)
- Draw North lines at every point mentioned
- Remember: bearings are measured clockwise from North
- Back bearing = bearing + 180° (or - 180° if over 360°)
- Use a protractor carefully - check which scale to use
- Draw diagrams to scale when asked
🧠 Problem-Solving Strategies
Problem-Solving
Bearings are measured clockwise from North (0 degrees). Always give bearings as 3 figures (e.g. 045 degrees, not 45 degrees). The bearing of B from A and the bearing of A from B differ by 180 degrees (back bearing). Draw a North line from every point used. Use angle facts and trigonometry to solve bearing problems.
Multi-Step Problem
Point B is 50 km from A on a bearing of 120 degrees. Point C is 80 km from B on a bearing of 210 degrees. Find the distance from A to C and the bearing of C from A.
Solution: Draw North lines at each point. At A: angle from North to AB = 120 degrees. At B: the bearing of 210 means the interior angle of triangle ABC at B = 180 - (210 - 180) - (180 - 120) = 180 - 30 - 60 = 90 degrees. Triangle ABC has a right angle at B. AC = sqrt(50 squared + 80 squared) = sqrt(2500 + 6400) = sqrt(8900) = 94.3 km. Bearing of C from A: angle CAB = arctan(80/50) = 58 degrees. Bearing = 120 + 58 = 178 degrees (approx).
⚠️ Common Errors
Watch Out!
1. Wrong: Measuring bearings anticlockwise from North Correct: Bearings are ALWAYS measured clockwise from North. An anticlockwise measurement is not a bearing.
2. Wrong: Writing a bearing as 45 degrees instead of 045 degrees Correct: All bearings must be given as 3-digit numbers: 005 degrees, 045 degrees, 180 degrees. This is standard navigational notation.
3. Wrong: Forgetting that the back bearing differs by 180 degrees Correct: If the bearing of B from A is 045 degrees, then the bearing of A from B is 045 + 180 = 225 degrees. Always add 180 for the return direction.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A ship sails from port P on a bearing of 075 degrees for 12 km to point Q. It then changes course and sails on a bearing of 150 degrees for 16 km to point R. (a) Calculate the distance PR. (b) Find the bearing of R from P. (c) The ship then returns directly to P. On what bearing does it sail?
(a) At Q, the angle between the two bearings: 150 - 75 = 75 degrees (interior angle). Using the cosine rule: PR squared = 12 squared + 16 squared - 2(12)(16)cos(75 degrees). PR squared = 144 + 256 - 384(0.2588) = 400 - 99.38 = 300.62. PR = 17.3 km.
(b) Using the sine rule: sin(angle P)/16 = sin(75)/17.3. sin(angle P) = 16 x 0.9659/17.3 = 0.8938. Angle P = 63.4 degrees. Bearing = 075 + 63.4 = 138 degrees (nearest degree).
(c) Bearing of P from R = bearing of R from P + 180 = 138 + 180 = 318 degrees.
Mark scheme: M1 cosine rule, A1 PR = 17.3, M1 sine rule, A1 bearing 138 degrees, M1 back bearing, A1 318 degrees
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two radar stations A and B are 100 km apart on a north-south line (A is north of B). A ship is on a bearing of 135 degrees from A and 060 degrees from B.
(a) How far is the ship from station A?
(b) The ship is sailing due East at 20 km/h. Will it pass closer to A or B in the next hour?
(c) A sailor says "The ship must be south-east of A because the bearing is 135 degrees." Is this correct? Explain.
Answers: (a) Angle at A = 135 - 180 = need to find the triangle. Bearing from A = 135 degrees, so angle between AB (due South) and AS = 180 - 135 = 45 degrees. Bearing from B = 060 degrees, so angle between BA (due North) and BS = 60 degrees. Triangle angle at S = 180 - 45 - 60 = 75 degrees. Using sine rule: AS/sin(60) = 100/sin(75). AS = 100 x sin(60)/sin(75) = 100 x 0.866/0.966 = 89.6 km. (b) Moving East from this position, the ship moves away from both stations horizontally, but distance change depends on geometry. It will generally pass closer to B since B is further south. (c) Yes — bearing 135 degrees from A means the ship is in the south-east direction from A. Bearing 060 degrees from B means it is north-east of B, consistent with being south-east of A.
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