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IB MYP Mathematics: Trigonometry and Pythagoras in MYP (MYP Mathematics) – Revision Notes

Condensed formulas, methods and a self-test on triangles, bearings, Pythagoras, trigonometry and the sine and cosine rules for IB MYP Mathematics.

Subject
Mathematics
Level
IB
Topic
Trigonometry and Pythagoras in MYP
Updated

Aligned to International Baccalaureate IB Middle Years Programme Mathematics (MYP Mathematics), From 2020, first assessment 2022. Official specification .

Syllabus page (what it covers and how it is assessed): IB Middle Years Programme Mathematics.

Syllabus points this page covers

MYP Mathematics

  • 3 Spatial reasoning

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Need help with this topic? Request a free trial class for IB Middle Years Programme Mathematics (MYP Mathematics).

Quick recall for the trigonometry rows of spatial reasoning in the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022. Full explanations are in the study guide. Test yourself with the practice questions, tick topics off on the checklist, and see the MYP mathematics hub. The free diagnostics help you choose what to revise first.

Labels follow the framework: standard is MYP 4-5 (standard), extended is MYP 4-5 (extended), enrichment is the framework’s enrichment list.

Key facts and formulas

Topic Result Label
Triangle angles Interior angles add to 180° Standard
Exterior angle Equals the sum of the two opposite interior angles Standard
Isosceles, equilateral Equal sides face equal angles; equilateral angles are 60° Standard
Side rules Longest side faces largest angle; two sides add to more than the third Standard
Bearing Clockwise from north, three figures; back bearing is plus or minus 180° Standard
Pythagoras a² + b² = c², c is the hypotenuse Standard
Ratios sin = O/H, cos = A/H, tan = O/A Standard
Converse If a² + b² = c² for the longest side, the triangle is right-angled Extended
Sine rule a/sin A = b/sin B = c/sin C Extended
Cosine rule a² = b² + c² − 2bc cos A Extended
3D distance √[(Δx)² + (Δy)² + (Δz)²] Enrichment
Area rule ½ab sin C Enrichment
Radians 180° = π rad Enrichment
Circle at origin x² + y² = r² Enrichment
Identities sin²θ + cos²θ = 1, tan θ = sin θ / cos θ Enrichment

Methods in steps

Choosing a tool (algorithmic thinking)

  1. Sketch and label. Mark right angles and north lines.
  2. Right angle present? Two sides known and a side wanted: Pythagoras. An angle involved: sin, cos or tan.
  3. No right angle? Side and its opposite angle known, plus one more value: sine rule. Two sides and the included angle, or all three sides: cosine rule.
  4. Check the result against the sketch. Round at the end.

Right-angled trigonometry

  1. Mark the angle you are using. Label opposite, adjacent, hypotenuse from it.
  2. Pick the ratio that links the two sides involved.
  3. Unknown on top of the fraction: multiply. Unknown on the bottom: divide.
  4. For an angle, use the inverse function.

Bearings

  1. Draw a north line at the starting point of each leg.
  2. Measure clockwise to the direction of travel.
  3. For the return direction, add or subtract 180°.
  4. Join legs with triangles and use the rules above.

Small worked reminders

  • Unknown on top. Angle 52°, adjacent 15 m, find the opposite: opposite = 15 tan 52° = 19.2 m.
  • Unknown on the bottom. Opposite 6 cm, angle 25°, find hypotenuse: 6 ÷ sin 25° = 14.2 cm.
  • Find an angle. Opposite 4.5, hypotenuse 12: sin θ = 4.5/12, so θ = 22.0°.
  • Back bearing. Bearing 072° out means 252° back.
  • Cosine rule order. Square and add the two sides, subtract 2bc cos A, then take the square root last.
  • Ambiguous case. sin B = 0.857 gives B = 59.0° and B = 121.0°. Test each with the angle sum.
  • Converse. 15, 20, 25: 225 + 400 = 625 = 25², so right-angled. 5, 6, 8: 25 + 36 = 61 but 8² = 64, so not.

Two fuller reminders

Isosceles triangle with the apex angle given. Equal sides 7 cm, apex 40°. Split it down the middle into two right-angled triangles. Each has hypotenuse 7 and an angle of 20° at the apex. Half the base is 7 sin 20°, so the base is 2 × 7 × sin 20° = 4.79 cm. The base angles are (180 − 40)/2 = 70° each.

Bearings with a right angle. A walker goes 5 km on bearing 090°, then 12 km on bearing 180°. The two legs meet at a right angle.

  • Distance from the start: √(25 + 144) = 13 km.
  • The angle below the east line is tan⁻¹(12/5) = 67.4°, so the bearing from the start is 090° + 67.4° = 157.4°.
  • The bearing of the start from the end is 157.4° + 180° = 337.4°.

Enrichment reminders

  • Area rule. Sides 6 and 7 enclosing 50°: ½ × 6 × 7 × sin 50° = 16.1.
  • Unit circle. At 210° the point is (cos 210°, sin 210°) = (−0.866, −0.5). Both are negative because the point is in the third quadrant.
  • Radians. 3π/2 rad = 270°. And 0.8 rad = 0.8 × 180/π = 45.8°.
  • Circle at the origin. A circle with centre (0, 0) through (−6, 8) has r² = 36 + 64 = 100, so the equation is x² + y² = 100.
  • Identity use. If cos θ = −0.6 and θ is obtuse, then sin θ = √(1 − 0.36) = 0.8 and tan θ = 0.8 ÷ (−0.6) = −4/3.
  • Proving an identity. Show that (1 − cos²θ) / sin θ = sin θ. Replace 1 − cos²θ with sin²θ, then divide: sin²θ / sin θ = sin θ.

Must-know distinctions

  • Pythagoras vs trigonometry. Pythagoras links sides only. Trigonometry links a side with an angle.
  • Sine rule vs cosine rule. The sine rule needs a matching side and angle pair. The cosine rule needs two sides and the included angle, or three sides.
  • Exact vs rounded. Keep full values in the calculator and round once, at the end.
  • Elevation vs depression. Elevation is upward from the horizontal. Depression is downward from the horizontal.
  • Bearing vs angle in the diagram. A bearing is clockwise from north. The triangle angle you need is often a different angle, found from parallel north lines.
  • Radians vs degrees. Multiply degrees by π/180 to get radians. The two modes must match the angle you enter.

Presenting solutions

  • Write the formula first, then substitute, then give the answer. This shows your reasoning (criterion C) and lets a reader follow where a slip happened.
  • Give a diagram for any bearings or elevation problem, with the given values marked on it.
  • State the degree of accuracy you use, for example 3 significant figures, and keep to it.
  • In a real-life task (criterion D), finish with a sentence that checks the answer against the situation: is a 240 m ladder reasonable? Does the distance fit the map?
  • In an investigation (criterion B), look for a pattern first, for example in Pythagorean triples such as 3, 4, 5 and 5, 12, 13, then test it on new cases before claiming a general rule.

Triangle property checks

  • Interior angles: add them. If they do not reach 180°, something is missing.
  • Exterior angle: use it as a shortcut instead of finding the third angle first.
  • Sides: before drawing, test the triangle inequality for the three lengths.
  • Isosceles: if two angles are equal, the sides opposite them are equal too.
  • Right-angled: if the hypotenuse is given, no other side can be longer than it.

Quick self-test

  1. Two angles of a triangle are 52° and 67°. Find the third.
  2. The bearing of B from A is 072°. Find the bearing of A from B.
  3. A right-angled triangle has legs 7 and 24. Find the hypotenuse.
  4. A right-angled triangle has hypotenuse 13 and one leg 5. Find the other leg.
  5. Find the side opposite 35° when the hypotenuse is 9.0 cm. Give 3 s.f.
  6. Find θ if tan θ = 0.75, to 1 d.p.
  7. (Extended) Is a triangle with sides 8, 15 and 17 right-angled? Show why.
  8. (Extended) In a triangle, a = 10, A = 30° and B = 100°. Find b to 3 s.f.
  9. (Extended) Two sides, 5 and 6, enclose an angle of 60°. Find the third side to 3 s.f.
  10. (Enrichment) Find the distance from (0, 0, 0) to (2, 3, 6), and write 100° in radians as a multiple of π.
  11. (Enrichment) If sin θ = 12/13 and θ is acute, find cos θ.

Answers

  1. 180 − 52 − 67 = 61°.
  2. 072° + 180° = 252°.
  3. √(49 + 576) = √625 = 25.
  4. √(169 − 25) = √144 = 12.
  5. 9.0 × sin 35° = 5.16 cm.
  6. θ = tan⁻¹(0.75) = 36.9°.
  7. 64 + 225 = 289 = 17², so yes, right-angled.
  8. b = 10 × sin 100° ÷ sin 30° = 19.7.
  9. √(25 + 36 − 2 × 5 × 6 × cos 60°) = √(61 − 30) = √31 = 5.57.
  10. √(4 + 9 + 36) = 7. 100° = 100π/180 = 5π/9.
  11. cos²θ = 1 − 144/169 = 25/169, so cos θ = 5/13.

Where marks are usually lost

  • Not drawing a sketch, so the wrong sides or angles are used.
  • Choosing sin when the sides given are opposite and adjacent.
  • Adding the squares when the unknown is a leg, rather than subtracting.
  • Measuring a bearing anticlockwise, or writing two figures instead of three.
  • Rounding in the middle of a multi-step problem.
  • Giving only one angle from an ambiguous case without checking the other.
  • Writing the answer with no units, or an angle with no degree sign.
  • Not justifying a converse test: show both sides of the comparison.
  • Writing a final answer that does not fit the context, such as a height longer than the ladder. Criterion D asks you to justify whether an answer makes sense.
  • Stating a conclusion without a clear line of reasoning, which affects criterion C.

Official syllabus

IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022, International Baccalaureate Organization: spatial reasoning framework, trigonometry rows.

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