Back-bearing rule: if bearing < 180°, add 180°. If bearing ≥ 180°, subtract 180°. So bearing of A from B is always within 0–360°.
Worked examples
WE 1Three-figure bearing and back-bearing
A boat is travelling in a direction 35° east of north. (a) Write this direction as a three-figure bearing. (b) Find the back-bearing of the boat’s starting point from its destination.
(a) “35° east of N” = 35° clockwise from N
three-figure bearing = 035°
(leading zero needed: must be 3 digits)
(b) back-bearing = bearing + 180° (since 035 < 180)
back-bearing = 035 + 180 = 215°
(a) 035° · (b) 215°
three digits is non-negotiable. Writing “35°” loses the mark even if the value is right. Always pad with leading zeros.
WE 2Right-angle journey — Pythagoras with bearings
A ship leaves port P and sails 12 km due east to point Q, then sails 5 km due south to point R. Find the direct distance from R back to P.
Sketch — turn at Q is from E to S, so 90°
PQ = 12 (east), QR = 5 (south)
PQR is a right triangle, right angle at Q
Apply Pythagoras
PR² = 12² + 5² = 144 + 25 = 169
PR = √169 = 13 km
PR = 13 km
(5, 12, 13) Pythagorean triple. “Due east then due south” gives a 90° turn — perfect for Pythagoras instead of cosine rule.
WE 3Cosine rule with bearings — find distance
A plane flies 50 km from airport A on a bearing of 060° to point B. From B it flies a further 30 km on a bearing of 120° to point C. Find the direct distance AC.
Find interior angle at B (back-bearing trick)
bearing of A from B = 060 + 180 = 240°
bearing of C from B = 120°
angle ABC = 240 − 120 = 120°
Apply cosine rule: two sides + included angle
AC² = 50² + 30² − 2(50)(30) cos(120°)
= 2500 + 900 − 3000(−0.5)
= 3400 + 1500 = 4900
AC = √4900 = 70 km
AC = 70 km
interior angle from two bearings at the same point = (back-bearing of leg 1) − (bearing of leg 2). Drawing the north line at B makes this obvious.
WE 4Right-angle journey — find both distance AND bearing
A drone leaves a depot D and flies 6 km due east, then 8 km due north to reach point F. Find: (a) the direct distance DF; (b) the bearing of F from D.
(a) Right-angle journey: Pythagoras
DF² = 6² + 8² = 36 + 64 = 100
DF = √100 = 10 km
(b) Bearing of F from D — F is NE of D
F is 6 km east and 8 km north of D
angle east-of-N = arctan(6/8) = arctan(0.75)
≈ 36.87°
bearing of F from D ≈ 037° (3 figures!)
(a) DF = 10 km · (b) bearing ≈ 037°
to find a bearing from coordinates: arctan(east/north). NE quadrant → bearing < 90°. Always confirm with a sketch.
WE 5Cosine + sine rule — distance and bearing of one ship from another
Two ships leave port P at the same time. Ship X sails 8 km on a bearing of 040°. Ship Y sails 5 km on a bearing of 100°. Find: (a) the distance XY; (b) the bearing of Y from X.
(a) angle XPY at port = 100° − 040° = 60°
cosine rule with SAS (8, 5, 60°)
XY² = 8² + 5² − 2(8)(5) cos(60°)
= 64 + 25 − 40 = 49
XY = 7 km
(b) Step 1 — angle PXY using sine rule
sin(PXY) / 5 = sin(60°) / 7
sin(PXY) = 5 sin(60°) / 7 ≈ 0.6186
angle PXY ≈ 38.21°
Step 2 — combine with back-bearing at X
bearing of P from X = 040 + 180 = 220°
Y lies west of XP at X (Y is south of P, X is NE of P)
bearing of Y from X = 220 − 38.21 ≈ 182°
(a) XY = 7 km · (b) bearing ≈ 182°
“find a bearing of one point from another” is a 3-step process: (1) cosine rule for the distance, (2) sine rule for the interior angle, (3) combine with the back-bearing at the vertex. A sketch determines whether you ADD or SUBTRACT the interior angle.
WE 6Construct a diagram — walker problem
A walker starts at A and walks 8 km due north to point B. From B, she walks 6 km on a bearing of 110° to point C. (a) Sketch the diagram, marking north lines at A and B. (b) Find the angle ABC. (c) Find the direct distance AC.
(a) sketch — A at the bottom, B 8 km north of A (above), C southeast of B
(b) angle ABC — use back-bearing at B
“A due N of B going S”: bearing of A from B = 180°
bearing of C from B = 110°
angle ABC = 180 − 110 = 70°
(c) cosine rule SAS: AB = 8, BC = 6, angle B = 70°
AC² = 8² + 6² − 2(8)(6) cos(70°)
= 64 + 36 − 96 × 0.342
= 100 − 32.83 ≈ 67.17
AC ≈ √67.17 ≈ 8.20 km (3 sf)
(b) angle ABC = 70° · (c) AC ≈ 8.20 km
“due north” gives a clean bearing of 000° (going up). The back-bearing at B is 180° (looking down at A). Subtract the next bearing to get the interior angle.
💡 Top tips
- Draw a north line at every vertex where a bearing is taken. Without it, you can’t see which angle is which.
- Use back-bearings to find interior angles: bearing AT a vertex looking BACK = (bearing of the path TO the vertex) ± 180°.
- Three figures always: 035°, not 35°. 005°, not 5°. 360° (or 000°) for due north.
- Pythagorean triples appear often: when bearings differ by exactly 90° (e.g. E then S), expect a 5-12-13 or 3-4-5 family.
- To find a bearing of one point from another: find the interior angle in the relevant triangle, then add/subtract from a known bearing at that vertex. Sketch tells you which.
âš Common mistakes
- Measuring anticlockwise or from another axis: bearings are CLOCKWISE from NORTH only. East is 090°, NOT 000° or 270°.
- Missing leading zeros: 35° is incorrect — write 035°. Marks are deducted in IB exam.
- Forgetting to draw the north line at the second point: bearings at B are measured from B‘s north line, not A‘s.
- Adding/subtracting bearings when you should use the back-bearing: bearing of A from B = (bearing of B from A) ± 180°, NOT the same as the original.
- Not deciding “add or subtract” with a sketch: when combining an interior angle with a bearing to get a new bearing, the side of the path determines the sign. Always sketch first.
That’s the end of Trigonometry. Across these four notes you have: SOH-CAH-TOA for right-angled triangles; sine rule, cosine rule, and area formula for any triangle; elevation and depression for vertical/horizontal applications; bearings for navigation. Every IB AI SL trigonometry question is a combination of these tools. Next chapter will likely be Statistics — descriptive statistics, frequency tables, measures of central tendency and spread.
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