Trigonometric Ratios
The fundamental trigonometric ratios (SOHCAHTOA) find unknown lengths and angles in right-angled triangles, and for the ESAT you must apply them in 2D and 3D scenarios and instantly recall the exact trig values for key angles like 30, 45, and 60 degrees.
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.