OxfordAQA International GCSE Mathematics: Pythagoras' theorem and right-angled trigonometry (9260)

This guide covers statement G19 in section 3.3.2 (Mensuration and calculation) of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. The Core content is Pythagoras’ theorem and the sine, cosine and tangent ratios in right-angled triangles in two-dimensional figures. The Extension content adds three-dimensional figures. All content can be assessed on the Extension papers, so Core learners should work through the 2D sections and Extension learners should do the whole guide. Scientific calculators are allowed on every paper.

Revise from the revision notes, then test yourself with the practice questions. The course hub and the printable checklist show where this unit sits, and the free 10-minute diagnostics help you find weak spots. For related number skills see Number.

What this unit covers

Spec reference	What you must be able to do	Level
G19	Know and use a² + b² = c²	Core
G19	Know sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent	Core
G19	Find lengths and angles in right-angled triangles in 2D figures	Core
G19	Apply the same ideas in 3D figures	Extension only

The specification says you must know the formulas, so learn them rather than hoping to look them up. Other units cover the sine rule, cosine rule and area of a triangle (non-right-angled triangles), and similar figures, arcs and sectors. This guide stays with right angles.

Labelling a right-angled triangle

Every right-angled triangle has three sides with fixed names.

• The hypotenuse is opposite the right angle. It is always the longest side.
• The opposite side faces the angle θ you are working with.
• The adjacent side touches θ and is not the hypotenuse.

The hypotenuse never changes, but opposite and adjacent swap if you move to the other acute angle. Always mark θ first, then label the three sides in the order hypotenuse, opposite, adjacent.

Pythagoras’ theorem

In a right-angled triangle with hypotenuse c and shorter sides a and b:

a² + b² = c²

The theorem links three lengths and works only in right-angled triangles.

• To find the hypotenuse, add the squares of the short sides, then take the square root.
• To find a short side, subtract the square of the known short side from the square of the hypotenuse, then take the square root.

The usual slip is adding when you should subtract. Check the answer: the hypotenuse must come out longer than both other sides.

Example 1: finding the hypotenuse. The short sides are 13.5 cm and 8.4 cm.

c² = 13.5² + 8.4²
   = 182.25 + 70.56
   = 252.81
c  = √252.81 = 15.9 cm

The answer is exactly 15.9 here because 15.9² = 252.81.

Example 2: finding a short side. The hypotenuse is 20.5 cm and one short side is 9.6 cm.

b² = 20.5² − 9.6²
   = 420.25 − 92.16
   = 328.09
b  = √328.09 = 18.1 cm (3 s.f.)

Example 3: a context. A support wire runs from the top of a vertical pole, 7.5 m above level ground, to a point on the ground 4.2 m from the foot of the pole. How long is the wire?

The pole and ground meet at a right angle, so the wire is the hypotenuse.

wire² = 7.5² + 4.2² = 56.25 + 17.64 = 73.89
wire  = √73.89 = 8.60 m (3 s.f.)

Draw the right-angled triangle for any word problem and write the lengths on it before you calculate.

The three trigonometric ratios

For an acute angle θ in a right-angled triangle:

Ratio	Formula
sin θ	opposite / hypotenuse
cos θ	adjacent / hypotenuse
tan θ	opposite / adjacent

A common memory aid is SOH CAH TOA. Make sure your calculator is working in degrees.

Choosing the ratio

Look at the side you know, the side you want and the angle. Two of the three sides are involved in every question. Match the pair to one ratio.

You have / want	Use
opposite and hypotenuse	sin
adjacent and hypotenuse	cos
opposite and adjacent	tan

Finding a length

Write the ratio as an equation, then solve it.

Example 4: unknown on top. A right-angled triangle has hypotenuse 14.6 cm. The angle θ is 33°. Find the side opposite θ.

Opposite and hypotenuse, so use sine.

sin 33° = x / 14.6
x = 14.6 × sin 33°
x = 7.95 cm (3 s.f.)

Example 5: unknown underneath. The side adjacent to a 57° angle is 7.9 cm. Find the hypotenuse h.

cos 57° = 7.9 / h
h = 7.9 / cos 57°
h = 14.5 cm (3 s.f.)

When the unknown is the denominator, rearrange before you use the calculator. Check that the hypotenuse is larger than 7.9.

Finding an angle

Use the inverse function (sin⁻¹, cos⁻¹ or tan⁻¹) on the ratio of the two known sides.

Example 6. Opposite 6.4 cm and adjacent 9.1 cm. Find θ.

tan θ = 6.4 / 9.1 = 0.7033
θ = tan⁻¹(0.7033) = 35.1° (1 d.p.)

Example 7. Opposite 5.2 cm and hypotenuse 8.3 cm. Find θ.

sin θ = 5.2 / 8.3 = 0.6265
θ = sin⁻¹(0.6265) = 38.8° (1 d.p.)

The third angle of a right-angled triangle is 90° − θ, because the angles add to 180°. Take the other acute angle this way rather than doing a second inverse calculation.

Combining the ideas in 2D figures

Many questions give a figure that is not itself a right-angled triangle. Look for a right angle in the figure, or add a line to make one. Two common moves:

• Split an isosceles triangle down the middle. The line from the apex to the midpoint of the base is perpendicular to the base and creates two matching right-angled triangles.
• Use two triangles in a row. Work out one length in the first triangle, then use it in the second.

Example 8: isosceles triangle. An isosceles triangle has equal sides of 9.5 cm and a base of 7 cm. Find its height and its base angle.

Split the triangle. Each half has hypotenuse 9.5 cm and base 3.5 cm.

h² = 9.5² − 3.5² = 90.25 − 12.25 = 78
h  = √78 = 8.83 cm (3 s.f.)

cos(base angle) = 3.5 / 9.5 = 0.3684
base angle = cos⁻¹(0.3684) = 68.4° (1 d.p.)

The apex angle is 180° − 2 × 68.4° = 43.2° (1 d.p.).

Notice the accuracy habit: use the unrounded value of h if you need it later. Rounded intermediate values give answers that are slightly out.

Extension only: three-dimensional figures

In a 3D figure you still use only right-angled triangles. The skill is to find the right one. Sketch the 3D shape, then draw the single flat triangle that holds the length or angle you need. Write its sides on a separate 2D sketch.

Lengths inside a cuboid

A cuboid has length l, width w and height h. The diagonal across the base is √(l² + w²). The space diagonal, from one corner to the opposite corner, forms a second right-angled triangle with that base diagonal and the vertical edge:

space diagonal² = l² + w² + h²

Example 9 (Extension only). A cuboid is 10 cm long, 6 cm wide and 7 cm high. Find (a) the space diagonal and (b) the angle it makes with the base.

(a) base diagonal² = 10² + 6² = 136
    space diagonal² = 136 + 7² = 185
    space diagonal = √185 = 13.6 cm (3 s.f.)

(b) The angle with the base is between the space diagonal
    and the base diagonal. Opposite = 7, adjacent = √136.
    tan θ = 7 / √136 = 0.6002
    θ = 31.0° (1 d.p.)

Angles between a line and a plane

The angle between a line and a plane is measured to the shadow of the line on the plane: drop a perpendicular from the top of the line to the plane, and the angle sits between the line and its shadow. In Example 9 the shadow of the space diagonal on the base is the base diagonal.

Pyramids

For a pyramid with its apex directly above the centre of a rectangular base, the vertical height meets the base at the centre. The centre is half-way along each base diagonal.

Example 10 (Extension only). A pyramid has a rectangular base 8 cm by 6 cm. The apex is 7 cm vertically above the centre of the base. Find (a) the length of a sloping edge, (b) the angle a sloping edge makes with the base, and (c) the angle the triangular face on an 8 cm base edge makes with the base.

(a) base diagonal = √(8² + 6²) = 10, so half = 5 cm
    edge² = 5² + 7² = 74
    edge  = √74 = 8.60 cm (3 s.f.)

(b) tan θ = 7 / 5, so θ = 54.5° (1 d.p.)

(c) The centre is 3 cm from the 8 cm edge (half of 6).
    tan φ = 7 / 3, so φ = 66.8° (1 d.p.)

In part (c) the angle between two planes is measured along a line at right angles to their meeting edge. That line runs from the midpoint of the base edge up to the apex.

Common errors

• Using the theorem on a triangle with no right angle.
• Adding the squares when finding a short side, or subtracting when finding the hypotenuse.
• Forgetting the square root at the end, so the answer is c² not c.
• Labelling opposite and adjacent from the wrong angle.
• Using tan⁻¹ on the wrong fraction by writing adjacent / opposite.
• Rearranging badly when the unknown is the denominator.
• Rounding in the middle of a multi-step problem.
• In 3D, using a slanting line where a perpendicular height is needed.

Next steps

Try the practice questions, then use the revision notes the week before your exam. Take a diagnostic when you want a quick check of where you stand.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.2 Mensuration and calculation, statement G19.