11.3 Pythagoras' Theorem & Trigonometry

Pythagoras' theorem is one of the most heavily tested ideas in the senior ICAS papers, and trigonometry is the signature skill of Papers I-J. Since personal calculators are banned and the on-screen calculator only appears when a question needs one, you must be fluent with exact square roots, the standard triples, and estimation — many Pythagoras questions are designed so the numbers come out exactly.

Pythagoras' Theorem

In any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the two shorter sides:

a² + b² = c²

where c is always the hypotenuse — the side opposite the right angle, and always the longest side. The theorem works ONLY for right-angled triangles.

Finding the Hypotenuse

Worked example: the shorter sides are 6 cm and 8 cm. Then c² = 6² + 8² = 36 + 64 = 100, so c = √100 = 10 cm. Add the squares, then square-root.

Finding a Shorter Side

Worked example: the hypotenuse is 13 cm and one shorter side is 12 cm. Then b² = c² − a² = 13² − 12² = 169 − 144 = 25, so b = √25 = 5 cm. Subtract the squares when you already have the hypotenuse. Adding them here (169 + 144 = 313) is the single most common mistake — always check which side is the hypotenuse before you start.

A sanity check catches most errors: the hypotenuse must be the longest side, but shorter than the sum of the other two sides.

Pythagorean Triples

A Pythagorean triple is a set of three whole numbers satisfying a² + b² = c². Knowing them turns a calculation question into a recognition question:

If a right-angled triangle has legs 9 and 12, spot the 3-4-5 triple scaled by 3: the hypotenuse is 15, no working needed. When the answer is not a whole number, leave it as an exact surd (e.g. √18 = 3√2) or round only at the final step.

Pythagoras in 2D and on the Coordinate Plane

Real problems hide the right angle: a ladder against a wall, the diagonal of a rectangle, the distance between two points. Always sketch and mark the right angle first.

Worked example: a rectangle is 24 m by 7 m. Its diagonal satisfies d² = 24² + 7² = 576 + 49 = 625, so d = 25 m (the 7-24-25 triple). On the coordinate plane, Pythagoras IS the distance formula from section 11.2: the distance between (1, 1) and (13, 6) is √(12² + 5²) = √(144 + 25) = √169 = 13 units.

The converse is also tested: if the sides of a triangle satisfy a² + b² = c², the triangle IS right-angled. So a triangle with sides 9, 12, 15 has a right angle (81 + 144 = 225), but one with sides 4, 5, 6 does not (16 + 25 = 41 ≠ 36).

Introducing Trigonometric Ratios (Papers I-J)

In a right-angled triangle, label the sides relative to the angle θ you are working with: the opposite (across from θ), the adjacent (next to θ, not the hypotenuse) and the hypotenuse. Then:

• sin θ = opposite / hypotenuse (SOH)
• cos θ = adjacent / hypotenuse (CAH)
• tan θ = opposite / adjacent (TOA)

Worked example — finding a side: in a right-angled triangle the hypotenuse is 10 cm and one angle is 30°. Find the side opposite that angle. sin 30° = opposite/10, so opposite = 10 × sin 30° = 10 × 0.5 = 5 cm. Values worth memorising: sin 30° = 0.5, cos 60° = 0.5, tan 45° = 1, sin 45° = cos 45° ≈ 0.707.

Worked example — finding an angle: the opposite side is 7 m and the adjacent side is 7 m. tan θ = 7/7 = 1, so θ = 45°. When the ratio is not a standard value, e.g. tan θ = 1.6, use the on-screen calculator's inverse function: θ = tan⁻¹(1.6) ≈ 58°. Round angles to the nearest degree and lengths to one or two decimal places — and only at the END, never mid-calculation.

Bearings

A bearing is an angle measured clockwise from north, always written with three figures: east is 090°, south is 180°, north-east is 045°. Bearings problems combine with Pythagoras or trigonometry.

Worked example: a boat sails 5 km due east, then 12 km due north. How far is it from its starting point, and on what bearing must it sail straight back? The legs are perpendicular, so the distance is √(5² + 12²) = √(25 + 144) = √169 = 13 km. For the return bearing, look at the triangle from the finish point: the start lies 5 km west and 12 km south, in the south-west quadrant. The angle between south and the return path satisfies tan = opposite/adjacent = 5/12 ≈ 0.417, giving about 23° west of south. Bearing is measured clockwise from north, so the return bearing is 180° + 23° = 203°. Always draw the north line at EVERY point and mark angles on your sketch.

The Sine and Cosine Rules (Papers I-J)

SOH CAH TOA works only in a right-angled triangle. Papers I and J extend to triangles with no right angle, where two further rules apply. Label each angle with a capital letter and the side opposite it with the matching lower-case letter: side a is opposite angle A, and so on.

The sine rule links each side to the sine of the angle opposite it:

a / sin A = b / sin B = c / sin C

Use it whenever you have a matched pair — a side together with the angle opposite it — plus one more side or angle.

Worked example: in triangle ABC, angle A = 40°, angle B = 75° and side a = 10 cm. Find side b. b / sin 75° = 10 / sin 40°, so b = 10 × sin 75° ÷ sin 40° ≈ 10 × 0.966 ÷ 0.643 ≈ 15.0 cm. Sanity check: B is the larger angle, so b must be the longer side, and 15.0 > 10. ✓

The cosine rule covers the two cases the sine rule cannot:

c² = a² + b² − 2ab cos C

Use it when you know two sides and the included angle (to find the third side), or all three sides (to find an angle, by rearranging to cos C = (a² + b² − c²) ÷ 2ab).

Worked example: two sides measure 7 cm and 9 cm with an included angle of 60°. Find the third side. c² = 7² + 9² − 2(7)(9) cos 60° = 49 + 81 − 126 × 0.5 = 130 − 63 = 67, so c = √67 ≈ 8.19 cm.

Notice that when C = 90°, cos 90° = 0 and the formula collapses to c² = a² + b² — the cosine rule contains Pythagoras as a special case, which is the quickest way to remember where the minus sign goes.

Choosing between them: if you can pair a side with its opposite angle, use the sine rule, because it needs less arithmetic; otherwise use the cosine rule. An angle found from the cosine rule is never ambiguous, whereas the sine rule can also admit an obtuse alternative — so if your answer looks too small for the diagram, check 180° minus that angle.

Trigonometry in 3D (Papers I-J)

Three-dimensional problems reduce to two right-angled triangles used in sequence. The classic example is the diagonal of a cuboid.

Worked example: a box measures 3 m by 4 m by 12 m. First find the diagonal of the base: √(3² + 4²) = √25 = 5 m. The space diagonal d then satisfies d² = 5² + 12² = 25 + 144 = 169, so d = 13 m. In general, the space diagonal of a cuboid is √(l² + w² + h²).

To find the angle a line makes with a plane, drop a perpendicular from the top of the line to the plane, then use trigonometry in the right-angled triangle formed by the line, its shadow on the plane, and the perpendicular. For the box above, the angle θ between the space diagonal and the base satisfies tan θ = height/base diagonal = 12/5 = 2.4, so θ = tan⁻¹(2.4) ≈ 67°. The strategy is always the same: identify the right-angled triangle containing the angle you want, extract it from the 3D figure, then apply SOH CAH TOA. If your answer gives an angle above 90° in a right-angled triangle, or a length longer than the space diagonal, something has gone wrong — estimate first and check.