Trigonometric Ratios
This topic covers the fundamental trigonometric ratios (SOHCAHTOA) used to find unknown lengths and angles in right-angled triangles. For the ESAT, you must apply these principles in 2D and 3D scenarios and instantly recall the exact trig values for key angles like 30°, 45°, and 60°.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- SOHCAHTOA is the mnemonic for the core definitions: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.
- These ratios only apply directly to right-angled triangles. For other triangles or 3D shapes, you must first identify or construct a relevant right-angled triangle.
- Pythagoras' theorem is often a required first step to find a missing side length before you can apply a trigonometric ratio.
- You are NOT expected to know or use the Sine Rule or Cosine Rule.
- Exact values for key angles must be memorised or quickly derived; answers will often be left in surd form.
Diagram
Formulae
sin(θ) = Opposite / Hypotenuse When you know or need to find the opposite side and hypotenuse relative to an angle.
cos(θ) = Adjacent / Hypotenuse When you know or need to find the adjacent side and hypotenuse relative to an angle.
tan(θ) = Opposite / Adjacent When you know or need to find the opposite and adjacent sides relative to an angle.
tan(θ) = sin(θ) / cos(θ) To find tan(θ) when sin(θ) and cos(θ) are known, or to relate the three ratios.
Definitions
- Hypotenuse
- The longest side of a right-angled triangle, always located opposite the right angle.
- Opposite
- The side of a right-angled triangle that is directly across from the angle of interest (θ).
- Adjacent
- The side of a right-angled triangle that is next to the angle of interest (θ), but is not the hypotenuse.
Worked example
A square-based pyramid has a base with side length 8 cm. The vertical height of the pyramid is 4 cm. Find the angle that a sloping face makes with the base.
- 1
First, draw a diagram of the pyramid.
Let the apex be V and the centre of the square base be O.
Let M be the midpoint of one of the base edges.
- 2
The angle required is the one inside the triangle VOM, specifically angle VMO.
- 3
This triangle VOM is a right-angled triangle, with the right angle at O.
- 4
The height VO is the side 'opposite' the angle.
We are given VO = 4 cm.
- 5
The length OM is the side 'adjacent' to the angle.
It is half the length of the square's side, so OM = 8 / 2 = 4 cm.
- 6
We can use the tangent ratio:
tan(θ) = Opposite / Adjacent - 7
Substitute the values:
tan(θ) = 4 / 4 = 1 - 8
Recall the exact trigonometric values.
The angle for which tan(θ) = 1 is 45°.
Answer: 45°
Common mistakes
- ×Arithmetic errors with surds, for example incorrectly calculating √(a2 + b2) or failing to simplify a result like 6/√(12).
- ×Misidentifying the Opposite, Adjacent, and Hypotenuse sides, especially on a rotated triangle or a complex 3D diagram.
- ×Forgetting to perform an initial Pythagoras calculation to find a necessary side length before applying SOHCAHTOA.
- ×Applying SOHCAHTOA to a triangle that is not right-angled. You must always isolate a right-angled triangle first.
No-calculator tips
- ✓Memorise two key triangles instead of the whole table of values. 1) An isosceles right-angled triangle with shorter sides of length 1 gives a hypotenuse of √(2) and is used for 45°. 2) An equilateral triangle of side length 2, bisected for height, creates a right-angled triangle with sides 1, √(3), and hypotenuse 2, used for 30° and 60°.
- ✓To simplify fractions with surds in the denominator (rationalising), multiply the top and bottom by that surd. For example, 8/√(2) = (8*√(2))/(√(2)*√(2)) = 8*√(2)/2 = 4*√(2).
- ✓When using Pythagoras, look for Pythagorean triples (like 3-4-5, 5-12-13) or their multiples to save calculation time.