Pearson Edexcel Edexcel 2021-style practice: In triangle ABC, AB = 9\,cm, AC = 7\,cm and the angle BAC = 50^. Work out the length of BC. (Higher tier, Paper 2, calculator.)

Two sides and the included angle are known, so use the cosine rule. a^2 = b^2 + c^2 - 2bccos A, where a = BC, b = 7, c = 9, A = 50^. BC^2 = 7^2 + 9^2 - 2(7)(9)cos 50^ = 49 + 81 - 126 × 0.643 = 130 - 81.0 = 49.0. BC = sqrt(49.0) ≈ 7.0\,cm. Markers award marks for the correct rule, substitution, and the final length. Using the sine rule (which needs an opposite pair) here is the common error.

EnglandMathsSyllabus dot point

How do you use Pythagoras' theorem and trigonometry to find sides and angles, including the sine and cosine rules?

Pythagoras' theorem in right-angled triangles, the trigonometric ratios sine, cosine and tangent, exact trig values, and the sine rule, cosine rule and area formula for any triangle (Higher tier).

A focused answer to the Edexcel GCSE Mathematics geometry content on Pythagoras and trigonometry, covering Pythagoras' theorem, the sine, cosine and tangent ratios, exact values, and the sine rule, cosine rule and triangle area formula at Higher tier.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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What this dot point is asking
Pythagoras' theorem
The trigonometric ratios
Exact trig values
The sine and cosine rules (Higher)

What this dot point is asking

Edexcel expects you to use Pythagoras' theorem and the three trigonometric ratios in right-angled triangles, to know exact trig values, and at Higher tier to use the sine rule, cosine rule and triangle area formula for any triangle. This is one of the most heavily examined Higher areas, so knowing which tool fits which situation is essential.

Pythagoras' theorem

Pythagoras' theorem connects the three sides of a right-angled triangle.

So a triangle with legs $5\,\text{cm}$ and $12\,\text{cm}$ has hypotenuse $\sqrt{5^2 + 12^2} = \sqrt{169} = 13\,\text{cm}$ . To find a shorter side when the hypotenuse is $13\,\text{cm}$ and one leg is $5\,\text{cm}$ : $\sqrt{13^2 - 5^2} = \sqrt{144} = 12\,\text{cm}$ . Pythagoras also gives the distance between two points and appears in 3D problems at Higher tier.

The trigonometric ratios

Trigonometry links an angle to the ratio of two sides in a right-angled triangle.

To find a side, label the sides relative to the angle, choose the ratio that uses the two sides you care about, and solve. To find an angle, use the inverse functions ( $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ). For example, if the opposite is $4$ and the hypotenuse is $8$ , then $\sin\theta = \tfrac{4}{8} = 0.5$ , so $\theta = \sin^{-1}(0.5) = 30^\circ$ .

Exact trig values

Some angles have exact values that you must know without a calculator, because they appear on the non-calculator paper.

The sine and cosine rules (Higher)

For triangles that are not right-angled, two rules extend trigonometry. Choosing between them is the key skill.

Choose and use the right rule

In triangle $PQR$ , $PQ = 8\,\text{cm}$ , angle $P = 40^\circ$ and angle $Q = 75^\circ$ . Find the length $QR$ .

step 1 Decide which rule fits

You have an angle ( $P$ ) and want the side opposite it ( $QR$ ), plus another angle-side pair is available, so use the sine rule.

step 2 Find the third angle if needed

$QR$ is opposite angle $P = 40^\circ$ . The side $PQ = 8$ is opposite angle $R$ . Find $R$ : $R = 180^\circ - 40^\circ - 75^\circ = 65^\circ$ .

step 3 Apply the sine rule

$\dfrac{QR}{\sin P} = \dfrac{PQ}{\sin R}$ , so $\dfrac{QR}{\sin 40^\circ} = \dfrac{8}{\sin 65^\circ}$ .

step 4 Solve

$QR = \dfrac{8 \sin 40^\circ}{\sin 65^\circ} = \dfrac{8 \times 0.643}{0.906} \approx 5.67\,\text{cm}$ .

Use the sine rule when you have a matching side-angle pair; use the cosine rule when you have two sides and the included angle (to find the third side) or all three sides (to find an angle). The area of any triangle is $\tfrac{1}{2}ab\sin C$ , using two sides and the angle between them. All three formulae are on the Edexcel formulae sheet.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20193 marksA right-angled triangle has the two shorter sides

6\,\text{cm}

and

8\,\text{cm}

. Work out the length of the hypotenuse. (Paper 1, non-calculator.)

Show worked answer →

Use Pythagoras' theorem: the square of the hypotenuse equals the sum of the squares of the other two sides.

$h^2 = 6^2 + 8^2 = 36 + 64 = 100$ .

$h = \sqrt{100} = 10\,\text{cm}$ .

Markers award a mark for squaring and adding, a mark for $100$ , and a mark for the square root giving $10\,\text{cm}$ . Subtracting the squares (used only when finding a shorter side) is the usual error.

Edexcel 20214 marksIn triangle

ABC

AB = 9\,\text{cm}

AC = 7\,\text{cm}

and the angle

BAC = 50^{\circ}

. Work out the length of

BC

. (Higher tier, Paper 2, calculator.)

Show worked answer →

Two sides and the included angle are known, so use the cosine rule.

$a^2 = b^2 + c^2 - 2bc\cos A$ , where $a = BC$ , $b = 7$ , $c = 9$ , $A = 50^\circ$ .

$BC^2 = 7^2 + 9^2 - 2(7)(9)\cos 50^\circ = 49 + 81 - 126 \times 0.643 = 130 - 81.0 = 49.0$ .

$BC = \sqrt{49.0} \approx 7.0\,\text{cm}$ .

Markers award marks for the correct rule, substitution, and the final length. Using the sine rule (which needs an opposite pair) here is the common error.

Related dot points

Sources & how we know this

Pearson Edexcel GCSE (9-1) Mathematics (1MA1) specification — Pearson Edexcel (2015)