OCR GCSE Mathematics: Pythagoras, right-angled trigonometry and exact trigonometric values (J560)

This guide follows OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), statements 10.05a, 10.05b and 10.05c. The 2D work and the exact values are in the Foundation tier and Higher tier content. Applying Pythagoras’ theorem and trigonometry in more complex figures, including 3D figures, is Higher tier only. The specification says that, unless it refers to calculator use, all content may be assessed on either the calculator or the non-calculator papers, so practise both ways. The sine rule and cosine rule are separate statements and are not covered here.

Use it with the revision notes and the practice questions. The OCR GCSE Mathematics hub and the printable checklist show where this unit sits, and the free 10-minute diagnostics find your weak spots first.

What this unit covers

Ref	You must be able to	Tier
10.05a	Know, derive and apply a² + b² = c² to find lengths in right-angled triangles in 2D figures	Foundation and Higher
10.05a	Apply Pythagoras’ theorem in more complex figures, including 3D figures	Higher tier only
10.05b	Know and apply sin θ, cos θ and tan θ to find angles and lengths in right-angled triangles in 2D figures	Foundation and Higher
10.05b	Apply the trigonometry of right-angled triangles in more complex figures, including 3D figures	Higher tier only
10.05c	Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60°, 90°, and of tan θ for θ = 0°, 30°, 45°, 60°	Foundation and Higher

Pythagoras’ theorem (10.05a)

In a right-angled triangle the longest side, opposite the right angle, is the hypotenuse. If its length is c and the other two sides are a and b:

a² + b² = c²

Deriving it

Take four copies of a right-angled triangle with short sides a and b and hypotenuse c. Arrange them round a square of side c, so that the four triangles sit inside a larger square of side (a + b). The large square has area (a + b)². It is also four triangles plus the tilted inner square:

(a + b)² = 4 × (½ab) + c²
a² + 2ab + b² = 2ab + c²
a² + b² = c²

Finding the hypotenuse

Square both short sides, add, then square root.

Example 1. Short sides 11 cm and 6 cm.

c² = 11² + 6² = 121 + 36 = 157
c = √157 = 12.5 cm (3 s.f.)

Finding a shorter side

Subtract. The hypotenuse is always the largest square.

Example 2. Hypotenuse 17.4 cm, one short side 9.3 cm.

a² = 17.4² − 9.3² = 302.76 − 86.49 = 216.27
a = √216.27 = 14.7 cm (3 s.f.)

Testing for a right angle

If the squares of the two shorter sides add to the square of the longest side, the triangle is right-angled.

Example 3. Is a triangle with sides 8, 15 and 17 right-angled?

8² + 15² = 64 + 225 = 289
17² = 289

The two results are equal, so the triangle is right-angled.

Pythagoras inside other shapes

Draw a perpendicular to split the shape into right-angled triangles.

Example 4. An isosceles triangle has equal sides of 25 cm and a base of 14 cm. Find its perpendicular height and area.

The height meets the base at its midpoint, so a short side is 7 cm.
h² = 25² − 7² = 625 − 49 = 576
h = 24 cm
Area = ½ × 14 × 24 = 168 cm²

On a coordinate grid, the distance between A(−2, 3) and B(5, 9) uses a horizontal step of 7 and a vertical step of 6:

AB² = 7² + 6² = 85
AB = √85 = 9.22 (3 s.f.)

Complex figures and 3D (Higher tier only)

Break the problem into right-angled triangles, one at a time, and use a length from one triangle in the next. Keep exact values such as √100 or √125 unrounded until the last step.

Example 5 (Higher tier only). A cuboid measures 8 cm by 6 cm by 5 cm. Find the length of the space diagonal, which runs from one corner to the opposite corner.

Diagonal of the 8 by 6 base: √(8² + 6²) = √100 = 10 cm
Space diagonal: √(10² + 5²) = √125 = 11.2 cm (3 s.f.)

Both steps combine into one result: d² = 8² + 6² + 5².

Right-angled trigonometry (10.05b)

Label the sides from the angle θ you are working with:

Side	Meaning
Hypotenuse	Opposite the right angle
Opposite	Across the triangle from θ
Adjacent	Next to θ, and not the hypotenuse

sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent

The ratios depend only on the angle, not on the size of the triangle. All right-angled triangles with the same angles are similar, so the ratios of their corresponding sides are equal (this links to similar shapes, 9.04c).

Choosing the ratio

Name the two sides involved: the one you know and the one you want. Then pick the ratio that uses exactly those two.

Finding a length

Example 6. Hypotenuse 12 m, angle 38°. Find the side opposite the angle.

sin 38° = opp / 12
opp = 12 × sin 38° = 7.39 m (3 s.f.)

Example 7. Adjacent side 6.5 cm, angle 52°. Find the opposite side.

tan 52° = opp / 6.5
opp = 6.5 × tan 52° = 8.32 cm (3 s.f.)

Example 8. The unknown is the hypotenuse. Opposite 5.4 cm, angle 28°.

sin 28° = 5.4 / h
h = 5.4 / sin 28° = 11.5 cm (3 s.f.)

When the unknown is on the bottom of the fraction, divide.

Finding an angle

Use the inverse function: sin⁻¹, cos⁻¹ or tan⁻¹.

Example 9. Opposite 9 cm, adjacent 14 cm.

tan θ = 9/14
θ = tan⁻¹(9/14) = 32.7° (1 d.p.)

Example 10. A 6.5 m ladder has its foot 2.5 m from a vertical wall. Find the angle the ladder makes with the ground, and how high it reaches.

cos θ = 2.5/6.5, so θ = cos⁻¹(2.5/6.5) = 67.4° (1 d.p.)
height² = 6.5² − 2.5² = 42.25 − 6.25 = 36
height = 6 m

The angle of elevation is measured upward from the horizontal. The angle of depression is measured downward from the horizontal.

More complex figures (Higher tier only)

Example 11 (Higher tier only). In triangle ABC, D lies on AC and BD is perpendicular to AC. AB = 14 cm, angle A = 35° and angle C = 50°. Find BC.

In triangle ABD: BD = 14 × sin 35° = 8.03 cm
In triangle BDC: sin 50° = BD / BC
BC = 8.03 / sin 50° = 10.5 cm (3 s.f.)

Keep the unrounded value of BD for the second step.

3D figures (Higher tier only)

To measure how steeply a line meets a plane, compare the line with its shadow on the plane, as if a light shone straight down. The line, the shadow and the vertical drop form a right-angled triangle.

Example 12 (Higher tier only). In the cuboid of Example 5, find the angle between the space diagonal and the 8 by 6 base.

The shadow of the diagonal on the base is the base diagonal, 10 cm.
The vertical edge is 5 cm.
tan θ = 5/10, so θ = tan⁻¹(0.5) = 26.6° (1 d.p.)

Example 13 (Higher tier only). A pyramid has a square base of side 10 cm. Its apex is 12 cm vertically above the centre of the base. Find the length of a sloping edge, and the angle between the sloping edge and the base.

Half of the base diagonal = ½ × √(10² + 10²) = 5√2 = 7.07 cm
Sloping edge² = (5√2)² + 12² = 50 + 144 = 194
Sloping edge = √194 = 13.9 cm (3 s.f.)
tan θ = 12 / (5√2), so θ = 59.5° (1 d.p.)

The angle between a sloping face and the base uses the perpendicular from the apex to the middle of a base edge instead: tan φ = 12/5, so φ = 67.4°. The slant height of that face is √(12² + 5²) = 13 cm.

Exact trigonometric values (10.05c)

Two triangles give every value you need.

Right-angled isosceles triangle with both short sides 1: the hypotenuse is √2 and the angles are 45°, 45°, 90°.

Equilateral triangle with side 2, cut in half: the base of each half is 1, the hypotenuse is 2, and the height is √(4 − 1) = √3. The angles are 30°, 60°, 90°.

Read the ratios from them:

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	√3/3	1	√3	not required

The specification asks for tan θ at 0°, 30°, 45° and 60° only. Note that √3/3 is the same as 1/√3, and √2/2 is the same as 1/√2. The cos row is the sin row reversed.

Example 14. Work out sin 60° × tan 30° ÷ cos 45°, giving an exact answer.

(√3/2) × (√3/3) = 3/6 = 1/2
(1/2) ÷ (√2/2) = 1/√2 = √2/2

Example 15. A right-angled triangle has a 30° angle and a hypotenuse of 8 cm. Find the side opposite the 30° angle.

opp = 8 × sin 30° = 8 × 1/2 = 4 cm

Example 16. A right-angled triangle has a 60° angle and an adjacent side of 5 cm. Find the opposite side exactly.

opp = 5 × tan 60° = 5√3 cm

Common errors

• Adding the squares when you should subtract. If you are finding a short side, you are subtracting.
• Taking the answer as c² and forgetting the final square root.
• Naming the opposite and adjacent sides from the wrong angle. Re-label whenever the angle changes.
• Dividing when you should multiply (or the other way round) when the unknown is on the bottom of the fraction.
• Writing sin 30° = √3/2. The two values 1/2 and √3/2 are swapped between sin 30° and sin 60°.
• Rounding a length part-way through a 2-step problem. Carry full values, then round once.
• In 3D, using a sloping edge as if it lay in the base. Draw the right-angled triangle separately and label it.

Next steps

Work through the practice questions, then use the revision notes for quick recall.

Official syllabus

OCR GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), first assessment in 2017, published by OCR. This unit covers statements 10.05a, 10.05b and 10.05c.