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How do you use Pythagoras theorem and trigonometry to find sides and angles?

Using Pythagoras theorem, the sine, cosine and tangent ratios in right-angled triangles, and the sine and cosine rules at Higher tier.

A focused answer to the AQA GCSE Mathematics geometry content on Pythagoras and trigonometry, covering Pythagoras theorem, the sine, cosine and tangent ratios in right-angled triangles, and the sine and cosine rules at Higher tier.

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  1. What this dot point is asking
  2. Pythagoras theorem
  3. Trigonometry in right-angled triangles
  4. The sine and cosine rules at Higher tier
  5. Putting it together with bearings
  6. Angles of elevation and depression
  7. Pythagoras and trigonometry in three dimensions

What this dot point is asking

AQA wants you to use Pythagoras theorem to find a side of a right-angled triangle, use the sine, cosine and tangent ratios to find sides and angles in right-angled triangles, and at Higher tier apply the sine rule and cosine rule to any triangle. This is a high-value strand: it appears in both papers and connects to bearings, 3D problems, area, and circle work.

Pythagoras theorem

To find the hypotenuse, add the squares of the two legs and square-root: legs 55 and 1212 give c=25+144=169=13c = \sqrt{25 + 144} = \sqrt{169} = 13. To find a shorter side, subtract: if the hypotenuse is 1010 and one leg is 66, the other is 10036=64=8\sqrt{100 - 36} = \sqrt{64} = 8. Pythagoras also gives the distance between two coordinate points and is used to find the slant height of cones and pyramids.

Trigonometry in right-angled triangles

Label the sides relative to the angle you are using: the hypotenuse is opposite the right angle, the opposite side faces the angle, and the adjacent is next to it. Then choose the ratio that links the side you know to the one you want.

You should also know the exact values: sin30=12\sin 30^\circ = \tfrac{1}{2}, cos60=12\cos 60^\circ = \tfrac{1}{2}, tan45=1\tan 45^\circ = 1, sin60=32\sin 60^\circ = \tfrac{\sqrt{3}}{2}, and cos30=32\cos 30^\circ = \tfrac{\sqrt{3}}{2}, which appear on non-calculator papers.

The sine and cosine rules at Higher tier

For triangles that are not right-angled, two extra rules cover every case.

Choose the rule from what you are given. Use the sine rule when you have a matching side and angle pair, such as one side, its opposite angle, and one more side or angle. 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). The area formula 12absinC\tfrac{1}{2}ab\sin C applies when you know two sides and the included angle.

Putting it together with bearings

Many Higher questions combine these tools with bearings. Sketch the triangle, mark the known sides and angles, decide whether it is right-angled (use SOH CAH TOA) or general (use the sine or cosine rule), and label clearly before substituting.

Angles of elevation and depression

A frequent application is the angle of elevation (looking up from the horizontal) or depression (looking down). These create right-angled triangles in vertical planes. To find the height of a tower from 30m30\,\text{m} away with an angle of elevation of 4040^\circ, the height is the opposite side and the distance is the adjacent, so tan40=h30\tan 40^\circ = \dfrac{h}{30}, giving h=30tan4025.2mh = 30 \tan 40^\circ \approx 25.2\,\text{m}. The angle of depression from the top of the tower to the same point equals the angle of elevation from the ground (alternate angles), a fact worth using to set up the triangle.

Pythagoras and trigonometry in three dimensions

Higher tier extends these tools into 3D, where you find lengths and angles inside solids such as cuboids and pyramids. The method is to identify a right-angled triangle within the solid, often using a diagonal of a face found by Pythagoras as one side of a larger triangle. For a cuboid, the space diagonal is found by applying Pythagoras twice, or directly by a2+b2+c2\sqrt{a^2 + b^2 + c^2}. The angle a diagonal makes with the base is then found with trigonometry on the triangle formed by the diagonal, its projection on the base, and the vertical. Drawing the relevant triangle separately, out of the 3D figure, makes these tractable.

Exam-style practice questions

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

AQA 20193 marksA right-angled triangle has legs of length 6cm6\,\text{cm} and 8cm8\,\text{cm}. Work out the length of the hypotenuse. (Foundation tier, Paper 1, non-calculator.)
Show worked answer →

By Pythagoras, the hypotenuse cc satisfies c2=62+82=36+64=100c^2 = 6^2 + 8^2 = 36 + 64 = 100.

So c=100=10cmc = \sqrt{100} = 10\,\text{cm}.

Markers award a mark for squaring and adding, a mark for the square root, and a mark for 10cm10\,\text{cm}. Subtracting the squares (instead of adding) when finding the hypotenuse is the classic error.

AQA 20224 marksIn triangle PQRPQR, PQ=7cmPQ = 7\,\text{cm}, QR=9cmQR = 9\,\text{cm} and the angle PQR=58PQR = 58^\circ. Work out the length of PRPR, giving your answer to 1 decimal place. (Higher tier, Paper 2, calculator.)
Show worked answer →

Two sides and the angle between them are known, so use the cosine rule a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A, with PRPR opposite the 5858^\circ angle.

PR2=72+922×7×9×cos58=49+81126×0.529913066.77=63.23PR^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos 58^\circ = 49 + 81 - 126 \times 0.5299 \approx 130 - 66.77 = 63.23.

So PR=63.237.9cmPR = \sqrt{63.23} \approx 7.9\,\text{cm}.

Markers reward choosing the cosine rule, correct substitution, and the rounded answer. Using the sine rule here (wrong information given) is the common mistake.

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