IB Maths AA SL Topic 3 — Geometry & Trig Paper 1 & 2 ~9 min read

Bearings & Constructions

A bearing is just a clever way to describe a direction — a 3-digit angle measured clockwise from north. Once you can read and draw bearings, every navigation problem becomes a regular trig problem with a fancy disguise.

📘 What you need to know

What is a bearing?

A bearing tells you which direction to head in. Imagine a compass with north at the top. Stand at one point, face north, then turn clockwise until you’re facing your destination. The angle you turned through (in degrees) is the bearing.

From North
Always start from the north line
Clockwise
Always turn clockwise (never anticlockwise)
3️⃣
3 Digits
Always 3-digit form, e.g. 045° not 45°
A bearing — measured clockwise from north
N E S W target 060° measured clockwise from north
Read this bearing as “060 degrees” or “060”. The direction shown is roughly east-north-east — 60° clockwise of due north.

The four cardinal directions

The compass has four cardinal points spaced 90° apart. Memorise these — they’re the anchor for all other bearings:

N
North
000°
E
East
090°
S
South
180°
W
West
270°

🧠 Memory trick — “Never Eat Soggy Waffles”

The cardinal directions go N, E, S, W clockwise. So as the bearing increases by 90°, you move clockwise: 000° → 090° → 180° → 270°. Some students remember the order as “Never Eat Soggy Waffles”.

Half-cardinals (NE, SE, SW, NW)

Half way between cardinals you’ve got the diagonals: NE = 045°, SE = 135°, SW = 225°, NW = 315°. They split each 90° quadrant in half.

Back bearings

If you know the bearing from A to B, you can work out the bearing from B back to A — that’s the back bearing. It’s always exactly 180° different.

Back bearing rule bearing < 180°  →  add 180°
bearing ≥ 180°  →  subtract 180°

Original < 180°

Add 180° to get the back bearing.

e.g. bearing 070° → back bearing

070° + 180° = 250°

Original ≥ 180°

Subtract 180° to get the back bearing.

e.g. bearing 230° → back bearing

230° − 180° = 050°

🤔 Why ± 180°?

If you walk in a straight line from A to B, the bearing from B back to A is exactly opposite — a half-turn (180°) different. Adding or subtracting 180° just keeps the answer in the valid 000°–360° range. Think of it as “flipping the arrow around”.

Solving bearings problems with trig

Once you’ve drawn the diagram with bearings marked, the problem usually reduces to a triangle question. The trick is finding the internal angle of the triangle — the angle inside the shape, not the bearing itself.

When a question gives you a bearing, draw a north line at every relevant point and measure the bearing from each. The angles between the bearing lines give you the internal angles of the triangle, which is what you’ll feed into the sine/cosine rule.

The standard 3-step approach

Bearings + trig method 1. Sketch the journey, marking north lines at each point
2. Use the bearings to find the internal angles of the triangle
3. Apply Pythagoras / SOH CAH TOA / sine rule / cosine rule
Typical bearings problem — finding internal angles
A N 050° B N 120° internal ∠B C Bearing of B from A = 050°, bearing of C from B = 120°. Find ∠ABC.
Quick angle calculation in the diagram above: the internal angle at B = 180° − 050° + 120° = wait, no — it’s actually (180 − 050) + 120 = 250°… but that’s reflex. The non-reflex internal angle = 360° − 250° = 110°. Always check which side of the angle you need!
When in doubt, mark every angle around point B that you can identify (using parallel north lines and angle facts). Then pick out the one inside your triangle.

Worked examples

WE 1

Express directions as bearings

Write each direction as a bearing in 3-digit form:

(a) due north  (b) north-east  (c) due south  (d) south-west

Step 1: Use the 4 cardinals + half-cardinals (a) North → 000° (b) North-East → halfway between 000° and 090° (b) 045° (c) South → 180° (d) South-West → halfway between 180° and 270° (d) 225° always 3 digits — write 045°, not 45°
WE 2

Find a back bearing

The bearing of a lighthouse from a ship is 075°. Find the bearing of the ship from the lighthouse.

Step 1: Original bearing is 075° < 180° Use the rule: add 180°. Step 2: Compute 075° + 180° = 255° Bearing of ship from lighthouse = 255° always check: original + back = 360 (or differ by 180)
WE 3

Distance using bearings + Pythagoras

A boat sails 8 km on a bearing of 000° (due north), then turns and sails 6 km on a bearing of 090° (due east). Find the direct distance from its starting point to its current position.

Step 1: Sketch the journey Two perpendicular legs (north then east) form a right-angled triangle. Step 2: Apply Pythagoras d2 = 82 + 62 = 64 + 36 = 100 Step 3: Square root d = 10 km classic 6-8-10 right triangle (3-4-5 scaled)
WE 4

Two-leg journey using cosine rule

A ship sails 15 km from A to B on a bearing of 040°, then 20 km from B to C on a bearing of 130°. Find the distance AC.

Step 1: Find the internal angle at B The two bearings differ by 130° − 040° = 90°. But we want the angle inside the triangle. Using the parallel north lines: angle ABC = 180° − 90° = 90°. turns out it’s a right angle — bonus! Step 2: Apply Pythagoras (or cosine rule with cos 90° = 0) AC2 = 152 + 202 = 225 + 400 = 625 Step 3: Square root AC = 25 km when the bearings differ by 90°, you get a right angle for free
WE 5

Find a bearing from given distances

From a point P, a tower T is 12 km due east, and a lighthouse L is 9 km due north of the tower. Find the bearing of L from P, to the nearest degree.

Step 1: Sketch — right-angled triangle PTL PT = 12 (east), TL = 9 (north), right angle at T. Step 2: The angle at P (from north line) is what we want From P, the north direction points up. Angle between PT (east) and PL (the line we want) is split by the north line. tan(angle from east) = 912 = 0.75 Angle from east = tan⁻¹(0.75) = 36.87° Step 3: Convert to bearing (clockwise from north) Bearing from P = 090° (east) − 36.87° = 053.13° Bearing of L from P ≈ 053° always state the bearing as a 3-digit number!

💡 Top tips

⚠ Common mistakes

Bearings questions are often the most “real-world” feeling problems on the IB paper — a great chance to show off your trig setup skills. Get the diagram right and the rest is just calculation.

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