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NotesMath AA SLTopic 3.3
Unit 3 · Geometry & Trigonometry · Topic 3.3

IB Math AA SL — Bearings & elevation

Topic 3.3 of IB Mathematics: Analysis and Approaches covers Bearings & elevation, which is part of Unit 3: Geometry & Trigonometry. Students explore key concepts including Elevation & depression, Bearings. A strong understanding of bearings & elevation is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Bearings & elevation

Key Idea: This topic turns a word problem about heights, distances and directions into a triangle you can solve. The exam pattern is always the same: read → draw the diagram → pick the rule (right-angled trig, or the sine/cosine rule).

📐 Elevation, depression & bearings

N = 000°, E = 090°, S = 180°, W = 270°. A direction like N40°E = 040°; S30°W = 180° + 30° = 210°; South-East = 135°. Back bearing = ±180° (A from B given B from A): add 180° if it's under 180°, subtract if it's 180° or more. Keep it between 000° and 360°.

🔺 Which rule? (Paper 1 by-hand, Paper 2 with GDC)


✏️ IB-style worked examples

IB-style question — elevation, find a height

From a point 65 m from the base of a tower, the angle of elevation to the top is 38°. Find the height of the tower.

Step by step:

  1. Height is opposite, 65 m is adjacent → use tan.

    tan⁡38∘=h65\tan 38^\circ = \frac{h}{65}tan38∘=65h​
  2. Solve for the height.

    h=65tan⁡38∘≈50.8 mh = 65\tan 38^\circ \approx 50.8\text{ m}h=65tan38∘≈50.8 m
Final answer:

About 50.8 m. (Add eye height if the angle is taken from a person's eye.)

IB-style question — depression, find a distance

From the top of a 55 m cliff, the angle of depression to a buoy is 24°. Find the buoy's distance from the base of the cliff.

Step by step:

  1. Depression (24°) equals the elevation from the buoy, so it sits inside the triangle.

    tan⁡24∘=55d\tan 24^\circ = \frac{55}{d}tan24∘=d55​
  2. Rearrange for the distance.

    d=55tan⁡24∘≈124 md = \frac{55}{\tan 24^\circ} \approx 124\text{ m}d=tan24∘55​≈124 m
Final answer:

About 124 m.

IB-style question — two observers, cosine rule for an angle

An object is seen from two points. The slant distances to it are 120 m and 150 m, and the two points are 90 m apart. Find the angle at the object.

Step by step:

  1. Three sides, angle wanted → cosine rule rearranged.

    cos⁡θ=1202+1502−9022(120)(150)\cos\theta = \frac{120^2 + 150^2 - 90^2}{2(120)(150)}cosθ=2(120)(150)1202+1502−902​
  2. Evaluate, then inverse-cosine.

    cos⁡θ=2880036000⇒θ≈36.9∘\cos\theta = \frac{28800}{36000} \Rightarrow \theta \approx 36.9^\circcosθ=3600028800​⇒θ≈36.9∘
Final answer:

About 36.9°.

IB-style question — bearing of one place, then its back bearing

A direction is given as 'North-West'. (a) Write it as a three-figure bearing. (b) If the bearing of B from A is 050°, find the bearing of A from B.

Step by step:

  1. (a) NW is 45° west of North (000°) — that is 45° back from a full turn of 360°.

    360∘−45∘=315∘360^\circ - 45^\circ = 315^\circ360∘−45∘=315∘
  2. (b) 050° is under 180°, so add 180° for the back bearing.

    050∘+180∘=230∘050^\circ + 180^\circ = 230^\circ050∘+180∘=230∘
Final answer:

(a) 315°. (b) 230°.

IB-style question — journey on two bearings, cosine rule

A ship sails 9 km on a bearing of 050°, then 6 km on a bearing of 110°. Find its direct distance from the start.

Step by step:

  1. Interior angle at the turn, from the North lines: 180° − (110° − 50°).

    ∠=180∘−60∘=120∘\angle = 180^\circ - 60^\circ = 120^\circ∠=180∘−60∘=120∘
  2. Two legs + included angle → cosine rule.

    d2=92+62−2(9)(6)cos⁡120∘=171d^2 = 9^2 + 6^2 - 2(9)(6)\cos 120^\circ = 171d2=92+62−2(9)(6)cos120∘=171
  3. Square-root for the distance.

    d=171≈13.1 kmd = \sqrt{171} \approx 13.1\text{ km}d=171​≈13.1 km
Final answer:

About 13.1 km.


Important: Bearings are measured clockwise from North, always with three digits. Sketch the North arrow first and sweep clockwise — never anticlockwise, never from South, never from the East line. A bearing drawn the wrong way sends every later angle and side astray.

Tap each card to reveal the answer.

Exam Tips

  • Always: read → draw the diagram → choose the rule. The sketch is where the marks start.
  • Bearings go clockwise from North and use three digits — 5° is 005°.
  • Depression equals the elevation back up; mark that equal angle inside your triangle.
  • Right-angled → SOH-CAH-TOA (often tan). Non-right → sine or cosine rule.
  • For journeys, find the interior angle at the turn from the North lines — it's rarely just the difference of the two bearings.

What you'll learn in Topic 3.3

  • 3.3.1 Elevation & depression
  • 3.3.2 Bearings
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 3.3 Bearings & elevation

3.3.1

Elevation & depression

Notes
3.3.2

Bearings

Notes

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Topic 3.3 Bearings & elevation forms a core part of Unit 3: Geometry & Trigonometry in IB Math AA SL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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