OCR OCR 2021-style practice: In a right-angled triangle the hypotenuse is 13 cm and the angle at the base is 35^. Work out the length of the side opposite the 35^ angle, to 1 decimal place. (Higher, Paper 4, calculator.)

The opposite side and the hypotenuse are linked by the sine ratio: sinθ = oppositehypotenuse. So sin 35^ = opp13, giving opp = 13 × sin 35^. sin 35^ ≈ 0.5736, so opp = 13 × 0.5736 ≈ 7.5 cm. Markers give a mark for choosing the sine ratio, a mark for rearranging to opp = 13 sin 35^, and a mark for 7.5 cm. Choosing the wrong ratio (cosine or tangent) is the usual error, which SOHCAHTOA prevents.

EnglandMathsSyllabus dot point

How do you use Pythagoras and the trigonometric ratios in right-angled triangles, and the sine and cosine rules in any triangle?

Use Pythagoras' theorem and the trigonometric ratios in right-angled triangles; and apply the sine rule, cosine rule and the area formula in any triangle (Higher tier).

A focused answer to the OCR GCSE Mathematics geometry content on Pythagoras and trigonometry, covering Pythagoras' theorem, the sine, cosine and tangent ratios, and the sine and cosine rules at Higher tier.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Pythagoras' theorem
The trigonometric ratios
The sine and cosine rules (Higher)
Why this matters

What this dot point is asking

OCR references G20, G21 and G23 cover Pythagoras' theorem and the trigonometric ratios in right-angled triangles, and at Higher tier the sine rule, cosine rule and the area formula for any triangle. Pythagoras and SOHCAHTOA must be recalled (they are not on the formulae sheet), while the sine and cosine rules are given on the Higher sheet. This is a high-value strand on every tier, appearing in lengths, angles, bearings and 3D problems.

Pythagoras' theorem

Pythagoras links the three sides of a right-angled triangle.

So a triangle with shorter sides $5$ and $12$ has hypotenuse $\sqrt{25 + 144} = \sqrt{169} = 13$ . To find a shorter side when the hypotenuse is $10$ and one side is $6$ , compute $\sqrt{100 - 36} = \sqrt{64} = 8$ . Pythagoras only works in right-angled triangles, and identifying the hypotenuse correctly is the first decision.

The trigonometric ratios

Trigonometry relates an angle to two sides.

To find a side, choose the ratio linking the angle, the known side and the unknown side, then rearrange. To find an angle, use the inverse function: if $\tan\theta = \tfrac{3}{4}$ , then $\theta = \tan^{-1}\!\left(\tfrac{3}{4}\right) \approx 36.9^\circ$ . Labelling the sides opposite, adjacent and hypotenuse before choosing the ratio prevents the most common mistake.

The sine and cosine rules (Higher)

For triangles without a right angle, two general rules apply.

Use the sine rule when you have a matching side-angle pair; use the cosine rule when you have two sides and the angle between them (to find the third side) or all three sides (to find an angle). These are on the OCR Higher formulae sheet, but selecting the right one is the skill the exam tests.

Use the cosine rule to find a side

A triangle has sides $b = 7$ cm and $c = 9$ cm with the included angle $A = 50^\circ$ . Find the third side $a$ .

step 1 Choose the rule

Two sides and the included angle are known, so use the cosine rule $a^2 = b^2 + c^2 - 2bc\cos A$ .

step 2 Substitute the values

$a^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos 50^\circ = 49 + 81 - 126\cos 50^\circ$ .

step 3 Evaluate

$\cos 50^\circ \approx 0.6428$ , so $a^2 = 130 - 126 \times 0.6428 = 130 - 80.99 = 49.01$ .

step 4 Square-root

$a = \sqrt{49.01} \approx 7.0$ cm.

Why this matters

Pythagoras and trigonometry turn geometry into calculation, and OCR uses them in ladders, ramps, bearings, isosceles triangles and 3D solids. The special-angle exact values ( $\sin 30^\circ = \tfrac{1}{2}$ , $\cos 30^\circ = \tfrac{\sqrt{3}}{2}$ , $\tan 45^\circ = 1$ ) are recalled facts that appear on the non-calculator paper. Choosing the correct tool, right-angled ratios or the general rules, and showing the substitution clearly is what secures the method marks.

Exam-style practice questions

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

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

6

cm and

8

cm. Work out the length of the hypotenuse. (Foundation, Paper 2, non-calculator.)

Show worked answer →

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

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

Take the square root: $c = \sqrt{100} = 10$ cm.

Markers award a mark for squaring and adding, a mark for $100$ , and a mark for $10$ cm. The standard error is forgetting the final square root, leaving the answer as $100$ .

OCR 20213 marksIn a right-angled triangle the hypotenuse is

13

cm and the angle at the base is

35^\circ

. Work out the length of the side opposite the

35^\circ

angle, to 1 decimal place. (Higher, Paper 4, calculator.)

Show worked answer →

The opposite side and the hypotenuse are linked by the sine ratio: $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ .

So $\sin 35^\circ = \dfrac{\text{opp}}{13}$ , giving opp $= 13 \times \sin 35^\circ$ .

$\sin 35^\circ \approx 0.5736$ , so opp $= 13 \times 0.5736 \approx 7.5$ cm.

Markers give a mark for choosing the sine ratio, a mark for rearranging to opp $= 13 \sin 35^\circ$ , and a mark for $7.5$ cm. Choosing the wrong ratio (cosine or tangent) is the usual error, which SOHCAHTOA prevents.

Related dot points

Sources & how we know this

OCR GCSE (9-1) Mathematics (J560) specification — OCR (2015)