Mathematics • Year 9 • Unit 3 • Lesson 17
True Bearings — Mixed Challenge
Bring together every idea from Lesson 17: quadrant-based conversions, reverse bearings, three-digit format, and tricky boundary cases. Spot a sign mistake, then attempt an open-ended bearing puzzle.
1. Mixed problems — pick the right rule
Each question targets a different feature of true bearings. Decide which formula or rule applies before you start writing. Show your working. 3 marks each
1.1 Convert each compass bearing to a true bearing (three-digit format): (a) N12°E, (b) S78°W, (c) N3°W.
1.2 Convert each true bearing to a compass bearing: (a) 095°, (b) 175°, (c) 355°. (Be careful with the boundary cases — 095 is just east of due east.)
1.3 Find the reverse bearing of (a) 360°, (b) 180°, (c) 090°. State the compass-form direction (or "due X") for both each original and its reverse.
1.4 A plane reports bearing N5°W. Convert to a true bearing. Then state how that bearing compares to "almost due north" — within how many degrees?
1.5 A boat sails on bearing 130°. After lunch the captain decides to head exactly perpendicular to her current direction, swinging to her LEFT (anti-clockwise). What is her new true bearing?
1.6 Two ships are at the same dock. Ship A leaves on bearing 080°. Ship B leaves on bearing 260°. (a) What is the relationship between these two bearings? (b) If they both travel at the same speed, what is the bearing of ship A from ship B after one hour? Justify your answer.
2. Find the mistake
Another student has tried to convert the compass bearing S35°W to a true bearing AND find its reverse. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — convert S35°W to true, then reverse:
Line 1: S35°W is in the SW quadrant.
Line 2: SW formula: true = 180 + θ.
Line 3: True = 180 + 35 = 215°.
Line 4: Reverse = 215 + 180 = 395°.
Line 5: So the reverse bearing is 395°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final reverse bearing.
Stuck? True bearings must stay between 000° and 360°. If you go over, subtract 360. (Alternative: since 215 is ≥ 180, you should subtract 180, not add.)3. Open-ended challenge — the bearing chain
This question has more than one valid answer — several different chains work. 4 marks
3.1 Design a chain of THREE consecutive true bearings (B1, B2, B3) — one bearing for each of three consecutive legs — such that ALL the following hold:
(i) B1 is in the NE quadrant (0° < B1 < 90°).
(ii) B2 is the reverse of B1.
(iii) B3 is the reverse of B2 (i.e. B3 is the reverse of the reverse of B1).
For your design:
(a) State your chosen B1.
(b) Compute B2 and B3 from the reverse-bearing rule.
(c) State the relationship between B1 and B3, and explain WHY in one sentence.
Bonus: Try a second value of B1 (still in the NE quadrant) and confirm the pattern.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Compass → true
(a) N12°E → NE: true = 12, pad: 012°.
(b) S78°W → SW: true = 180 + 78 = 258°.
(c) N3°W → NW: true = 360 − 3 = 357°.
1.2 — True → compass
(a) 095° in SE (just past 090): θ = 180 − 95 = 85. Compass: S85°E.
(b) 175° in SE: θ = 180 − 175 = 5. Compass: S5°E.
(c) 355° in NW: θ = 360 − 355 = 5. Compass: N5°W.
1.3 — Reverse of cardinals
(a) 360° = due north; reverse = 360 − 180 = 180° = due south.
(b) 180° = due south; reverse = 180 − 180 = 000° or equivalently 360° = due north.
(c) 090° = due east; reverse = 090 + 180 = 270° = due west.
1.4 — N5°W to true
NW quadrant: true = 360 − 5 = 355°. This is only 5° away from 360 (= due north), so it's within 5° of due north — almost exactly north, slightly tilted west.
1.5 — Boat on 130°, turn 90° left
Turning LEFT means anti-clockwise, i.e. SUBTRACT 90°: new bearing = 130 − 90 = 040°.
If she had turned right (clockwise), she'd be on 130 + 90 = 220°.
1.6 — Ships on 080° and 260°
(a) 260 − 080 = 180 → the two bearings are reverse bearings of each other. Ships A and B sail in exactly opposite directions.
(b) After one hour they are at equal distance on opposite sides of the dock on the same straight line. From ship B's perspective, ship A is back along the line ship B came from and continuing through the dock — that's ship A's outbound direction, which is bearing 080°.
Equivalently: ship A from ship B is in the direction of ship A's outward bearing — because A is "ahead in the direction A is going" relative to B's position behind the dock.
2 — Find the mistake (S35°W reverse)
(a) The mistake is on Line 4.
(b) True bearings must be in the range 000-360. After computing 215 + 180 = 395, the student forgot to subtract 360 to bring it back into range. (Equivalently, because 215 is ≥ 180, the simpler rule is to SUBTRACT 180, not add — both give the same final answer.)
(c) Corrected working:
S35°W in SW quadrant: true = 180 + 35 = 215°. ✓
Reverse: 215 ≥ 180, so subtract 180: 215 − 180 = 035°.
(Or: 215 + 180 = 395 → 395 − 360 = 035° — same answer.)
Sanity check: the original direction is SW; reversing should put us in NE. 035° is in NE quadrant. ✓
3 — Reversing twice (sample solution)
Design 1: B1 = 050° (in NE).
B2 = reverse(050) = 050 + 180 = 230° (in SW).
B3 = reverse(230) = 230 − 180 = 050°.
Relationship: B3 = B1. Why: reversing flips the direction (180°); reversing again flips it back, so two reverses = no change. ± 180° twice gives ± 360°, which equals 0° modulo 360.
Design 2 (bonus): B1 = 075° → B2 = 255° → B3 = 075°. Pattern confirmed. ✓
Marking: 1 mark for a valid B1 in NE; 1 for correct B2; 1 for correct B3 = B1; 1 for the "two reverses = no change" justification (or for a confirming second example).