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TRIGONOMETRY LAB
Learn it by playing it
Four short missions on two live triangles. Steer the legs, the angle and the size, watch the sides and the ratios react, and figure out the rules. When a mission glows green, read why it worked.
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Mission 1 of 4
Three, four, five
Two dials set the legs of a right-angled triangle; the hypotenuse stretches to match. Make the hypotenuse exactly 5 - with both legs whole numbers bigger than zero.
a² + b² = 4 + 4 = 8 · hyp = √8 = 2.83
Before you touch the dials: a right-angled triangle has legs 3 and 4. How long is the hypotenuse?
Currently 2^2 + 2^2 = 8 and the hypotenuse is 2.83 - target 5 with whole-number legs.
Study notes
Pythagoras' theorem C6.1 / E6.1
In a right-angled triangle the hypotenuse is the side opposite the right angle - always the longest side. Pythagoras' theorem: (hypotenuse)² = a² + b² for the other two sides a and b.
Finding the hypotenuse: square both legs, add, then take the square root. Example: legs 6 and 8 give 36 + 64 = 100, so the hypotenuse is 10.
Finding a leg: rearrange to leg² = (hypotenuse)² - (other leg)². Example: hypotenuse 13, one leg 5, so the other leg is √(169 - 25) = √144 = 12.
The theorem works only on right-angled triangles, and it connects LENGTHS - never mix it with angles.
Round answers the way the question asks: with legs 5 and 5 the hypotenuse is √50 ≈ 7.1 to 1 decimal place - an approximation, not an exact value.
Some whole-number triples are worth recognising on sight: 3-4-5 (and its multiples 6-8-10, 9-12-15), 5-12-13, 8-15-17.
You can also run the theorem backwards as a TEST: if the longest side squared equals the sum of the other two squares, the triangle is right-angled. 4-5-6 fails the test (36 ≠41), so it is not right-angled.
The three ratios: sin, cos, tan C6.2 / E6.2
Label the sides from the angle θ you are working with: the hypotenuse is opposite the right angle, the opposite is across from θ, and the adjacent is the remaining side next to θ.
sin θ = opposite / hypotenuse, cos θ = adjacent / hypotenuse, tan θ = opposite / adjacent. The mnemonic SOH CAH TOA keeps the three pairings straight.
The ratios belong to the ANGLE, not the triangle: scale a triangle up or down and every ratio stays the same. That is why one measured angle plus one measured side is enough to find all the others.
Angles of elevation and depression (E6.2) are the same idea in disguise: both are measured from the horizontal line, and the angle of elevation to a point equals the angle of depression back down from it.
Worked example: the angle is 35° and the hypotenuse is 10 m. The opposite side uses sin: opposite = 10 x sin 35° ≈ 5.7 m.
Worked example: the angle is 30° and the adjacent side is 8 cm. The opposite uses tan: opposite = 8 x tan 30° ≈ 4.6 cm.
Worked example: opposite 5, adjacent 12. tan θ = 5/12 ≈ 0.4167, and the inverse ratio gives θ ≈ 22.6°.
Answers to angles are given to 1 decimal place; keep at least 4 figures of the ratio in your working and round only at the end.
Exact values to know (Extended) E6.3
These values are exact, not calculator roundings - exams can ask for them without a calculator.
cos runs the same list backwards: 0° → 1, 30° → √3/2, 45° → 1/√2, 60° → 1/2, 90° → 0. Notice sin 30° = cos 60°: co-angles share values.
tan: 0° → 0, 30° → 1/√3, 45° → 1, 60° → √3. tan is undefined at 90° because the adjacent side has shrunk to zero.
Where they come from: half of an equilateral triangle gives the 30-60-90 values, and half of a square (a 45-45-90 triangle) gives the 45° values - so each one can be re-derived if forgotten.
Graphs of sin, cos and tan (Extended) E6.4
y = sin x is a wave between -1 and 1 with period 360°: it starts at 0, peaks at 90°, crosses zero at 180°, bottoms at 270°.
y = cos x is the same wave shifted 90° to the left: it starts at 1 and first crosses zero at 90°.
y = tan x repeats every 180° and shoots off to infinity near 90° (and 270°) - the curve never touches those vertical asymptotes.
Solving sin x = 0.5 for 0° ≤ x ≤ 360°: the calculator gives 30°, and the symmetry of the wave gives the second solution 180° - 30° = 150°.
The same symmetry pattern solves cos and tan equations: cos x = k has solutions θ and 360° - θ; tan x = k has solutions θ and 180° + θ within one full turn.
Any triangle: sine rule, cosine rule, area (Extended) E6.5 / E6.6
When the triangle has no right angle, Pythagoras and SOH CAH TOA do not apply - use the sine and cosine rules instead. Label sides a, b, c opposite angles A, B, C.
Sine rule: a / sin A = b / sin B = c / sin C. Use it when you know a matching pair (an angle and its opposite side) plus one more piece.
Cosine rule: a² = b² + c² - 2bc cos A. Use it when you know two sides and the angle BETWEEN them (to find the third side), or all three sides (to find an angle). Example: b = 5, c = 7, A = 60° gives a² = 25 + 49 - 35 = 39, so a ≈ 6.2.
Area of any triangle: ½ab sin C - half the product of two sides times the sine of the angle between them. Example: sides 8 and 10 with 30° between them give ½ x 8 x 10 x 0.5 = 20.
Three-dimensional problems (E6.6) reduce to flat right-angled triangles: for a cuboid 3 x 4 x 12 the space diagonal is √(9 + 16 + 144) = √169 = 13 - Pythagoras applied twice. Draw and label the triangle you are using before calculating.
Choosing the tool: right angle present → SOH CAH TOA; matching angle-side pair → sine rule; two sides plus included angle (or three sides) → cosine rule; area needed → ½ab sin C.
EduTribe study notes - original material aligned to the Cambridge IGCSE Mathematics 0580 syllabus (2025-2027). Automated checks completed; not yet reviewed by an educator.