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

Use Pythagoras' theorem and the trigonometric ratios in right-angled triangles, and the sine rule, cosine rule and area formula in any triangle (Higher tier), to find missing lengths and angles.

A focused answer to the WJEC 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 area formula for any triangle at Higher tier.

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

What this dot point is asking

WJEC asks you to use Pythagoras' theorem and the trigonometric ratios in right-angled triangles to find missing sides and angles, and at Higher tier to use the sine rule, the cosine rule and the area formula in any triangle. The key judgement is choosing the right tool: Pythagoras when only sides are involved, the trig ratios when a right angle and one angle are involved, and the sine or cosine rule when the triangle has no right angle. It is one of the most heavily assessed geometry strands and appears on both components.

Pythagoras' theorem

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

So a triangle with shorter sides 55 and 1212 has hypotenuse 52+122=169=13\sqrt{5^2 + 12^2} = \sqrt{169} = 13. The same idea finds the distance between two points on a grid by treating the horizontal and vertical gaps as the two shorter sides.

The trigonometric ratios

The ratios link an angle to two of the sides in a right-angled triangle.

Label the sides relative to the angle you are using: the hypotenuse (H) is opposite the right angle, the opposite (O) faces your angle, and the adjacent (A) is next to it. Then SOH CAH TOA gives sinθ=OH\sin\theta = \tfrac{O}{H}, cosθ=AH\cos\theta = \tfrac{A}{H} and tanθ=OA\tan\theta = \tfrac{O}{A}. To find a side, choose the ratio with the two sides you have, substitute and solve. To find an angle, use the inverse (sin1\sin^{-1}, cos1\cos^{-1}, tan1\tan^{-1}) on the ratio of the two known sides.

Choosing trig or Pythagoras

The given information tells you which tool to reach for.

If the question gives only sides and a right angle, use Pythagoras; if it gives an angle and asks for a side (or two sides and asks for an angle) in a right-angled triangle, use the ratios.

The sine and cosine rules (Higher)

When a triangle has no right angle, two rules from the formulae list take over.

A quick decision: if a side is paired with its opposite angle, the sine rule is usually fastest; if not, reach for the cosine rule. Both are printed in the WJEC formulae list, but you must know when each applies.

Why this matters

Pythagoras and trigonometry are among the most frequently examined geometry topics and combine with mensuration, bearings and vectors on harder questions. The marks reward selecting the correct method, labelling sides accurately and rounding sensibly, often to a stated number of decimal places or significant figures. At Higher tier the sine and cosine rules extend the same reasoning to any triangle, and recognising which rule fits the given information is the marked judgement that separates full marks from a stalled attempt.

Exam-style practice questions

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

WJEC 20183 marksA right-angled triangle has the two shorter sides 66 cm and 88 cm. Work out the length of the hypotenuse. (Foundation and Higher, Unit 1, non-calculator.)
Show worked answer →

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 100100 and a mark for the square root 1010 cm. Subtracting the squares (used only when finding a shorter side) is the common error here.

WJEC 20224 marksIn triangle ABC, AB=9AB = 9 cm, AC=7AC = 7 cm and the angle BAC =64= 64^\circ. Work out the length of BC, giving your answer to 1 decimal place. (Higher, Unit 2, calculator.)
Show worked answer →

Two sides and the included angle point to the cosine rule: a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A, with a=BCa = BC opposite the 6464^\circ angle.

BC2=92+722×9×7×cos64=81+49126cos64BC^2 = 9^2 + 7^2 - 2\times 9\times 7\times \cos 64^\circ = 81 + 49 - 126\cos 64^\circ.

cos64=0.4384\cos 64^\circ = 0.4384, so BC2=13055.23=74.77BC^2 = 130 - 55.23 = 74.77, giving BC=74.77=8.6BC = \sqrt{74.77} = 8.6 cm to 1 decimal place.

Markers give marks for the correct rule, the substitution, the value of BC2BC^2 and the final length. Forgetting that the cosine rule needs the included angle is the usual slip.

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