Solving problems using the sine and cosine rule - Higher
Click to explore updated revision resources for GCSE Maths: Higher - 2D and 3D trigonometry problems, with step-by-step slideshows, quizzes, practice exam questions, and more!
To solve problems involving non-right-angled triangles, the correct rule must first be chosen.
The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known, eg a, b and C
- an angle when three sides are known, eg a, b and c
The sine rule can be used in any triangle to calculate:
- a side when two angles and an opposite side are known, eg a, A and B
- an angle when two sides and an opposite angle are known, eg a, A and b
Example
Ship A leaves port P and travels on a A direction measured in degrees clockwise from North. Due North is 000, due East is 090, due South is 180 and due West is 270. of 200°. A second ship B leaves the same port P at the same time and travels on a bearing of 165°. After half an hour ship A has travelled 5.2 km and ship B has travelled 5.8 km. How far apart are the ships after half an hour? Give the answer to three significant figures.
Remember a bearing is an angle measured clockwise from north. The angle APB = \(200 - 165 = 35^\circ\).
Two sides and the angle in between are known. Calculate the length AB.
Use the cosine rule.
\(a^2 = b^2 + c^2 - 2bc \cos{A}\)
\(\text{AB}^2 = 5.2^2 + 5.8^2-2 \times 5.2 \times 5.8 \cos{35}\)
\(\text{AB}^2 = 11.26874869\)
AB = 3.36 km
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- Angles, lines and polygons - AQA
- Loci and constructions - AQA
- 2-dimensional shapes - AQA
- 3-dimensional shapes - AQA
- Circles, sectors and arcs - AQA
- Circle theorems - Higher - AQA
- Transformations - AQA
- Pythagoras' theorem - AQA
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