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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 Eduqas GCSE Mathematics geometry content on Pythagoras and trigonometry, covering Pythagoras' theorem, the sine cosine and tangent ratios in right-angled triangles, and the sine rule, cosine rule and triangle area formula at Higher tier.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Pythagoras' theorem
  3. The trigonometric ratios
  4. The sine and cosine rules (Higher)
  5. Why this matters

What this dot point is asking

The Eduqas geometry content asks you to use Pythagoras' theorem and the three trigonometric ratios in right-angled triangles at both tiers, and at Higher tier to apply the sine rule, cosine rule and the triangle area formula in any triangle. Pythagoras finds a missing side from the other two; trigonometry links sides and angles. These are among the most useful tools on the paper, feeding into bearings, area, vectors and 3D problems. The right-angled work is core; the sine and cosine rules are reliable Higher-tier multi-mark questions, and they are printed on the Eduqas formulae list.

Pythagoras' theorem

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

So to find a shorter side when the hypotenuse is 1313 and one side is 55, compute 13252=16925=144=12\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12. The direction of the calculation matters: add the squares for the hypotenuse, subtract for a shorter side. On the non-calculator Component 1, answers are often left in surd form, such as 20=25\sqrt{20} = 2\sqrt{5}.

The trigonometric ratios

Trigonometry connects an angle to the ratio of two sides, using SOH CAH TOA.

To find a missing side, choose the ratio that uses the side you know and the side you want, then rearrange. To find a missing angle, use the inverse function: if tanθ=34\tan\theta = \tfrac{3}{4}, then θ=tan1(34)36.9\theta = \tan^{-1}\left(\tfrac{3}{4}\right) \approx 36.9^\circ. Choosing the correct ratio from the labelled sides is the key decision.

The sine and cosine rules (Higher)

For triangles without a right angle, two rules extend trigonometry, both on the formulae list.

The choice depends on what you know: use the sine rule when you have a matching side-angle pair, and 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).

Why this matters

Pythagoras and trigonometry turn geometric diagrams into solvable equations, which is why they recur in bearings, area, vectors and 3D problems across both components. Eduqas often expects exact surd answers on the non-calculator paper and accurate calculator work on the other, and the special-angle values (sin30=12\sin 30^\circ = \tfrac{1}{2}, tan45=1\tan 45^\circ = 1, cos60=12\cos 60^\circ = \tfrac{1}{2}) are worth knowing because they appear without a calculator.

Exam-style practice questions

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

Eduqas 20193 marksA right-angled triangle has the two shorter sides of length 6 cm and 8 cm. Work out the length of the hypotenuse. (Foundation, Component 2, calculator.)
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Use Pythagoras' theorem a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse.

c2=62+82=36+64=100c^2 = 6^2 + 8^2 = 36 + 64 = 100.

c=100=10c = \sqrt{100} = 10 cm.

Markers award a mark for squaring and adding, a mark for the method, and a mark for the answer 10 cm. Forgetting to square root at the end (leaving the answer as 100) is the standard slip.

Eduqas 20224 marksIn triangle ABC, angle A=40A = 40^\circ, side b=9b = 9 cm and side c=12c = 12 cm. Work out the length of side aa using the cosine rule. Give your answer to 1 decimal place. (Higher, Component 2, calculator.)
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The cosine rule is a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A.

Substitute: a2=92+1222(9)(12)cos40=81+144216×0.766=225165.5=59.5a^2 = 9^2 + 12^2 - 2(9)(12)\cos 40^\circ = 81 + 144 - 216 \times 0.766 = 225 - 165.5 = 59.5.

So a=59.57.7a = \sqrt{59.5} \approx 7.7 cm.

Markers give marks for the correct rule, for substituting the values, for evaluating the cosine term, and for the final length. Using the sine rule (which needs an opposite side and angle pair) instead of the cosine rule is the common error here.

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