Trigonometry in IGCSE Maths
Right-angled triangle trigonometry (SOHCAHTOA), the sine rule, the cosine rule and the area formula — with clear diagrams, worked examples and an exam-style question. Learn which method to use and how to set out your working to earn full marks.
0580 · 6.1 / E6.2Labelling a right-angled triangle
Before you can use trigonometry you need to label the three sides of a right-angled triangle relative to the angle you are working with (here \(\theta\)):
The hypotenuse is always the longest side, opposite the right angle. The opposite faces the angle \(\theta\). The adjacent sits between \(\theta\) and the right angle.
The mnemonic SOH·CAH·TOA gives the three ratios:
\[\sin\theta=\frac{\text{O}}{\text{H}}\qquad\cos\theta=\frac{\text{A}}{\text{H}}\qquad\tan\theta=\frac{\text{O}}{\text{A}}\]where O = opposite, A = adjacent and H = hypotenuse.
SOHCAHTOA: finding a side
Decide which two sides are involved — the one you know and the one you want. Pick the ratio that uses those two sides, then rearrange to make the unknown the subject.
Find the length \(x\).
We know the hypotenuse (\(12\) cm) and want the opposite (\(x\)). Opposite and hypotenuse → use sine:
\[\sin 35° = \frac{x}{12}\] \[x = 12 \times \sin 35° = 6.88\text{ cm (3 s.f.)}\]SOHCAHTOA: finding an angle
When you know two sides and want the angle, choose the ratio that uses those two sides, then apply the inverse function (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\)) on your calculator.
Find the angle \(\theta\).
We know the opposite (\(7\)) and the adjacent (\(10\)). Opposite and adjacent → use tangent:
\[\tan\theta = \frac{7}{10} = 0.7\] \[\theta = \tan^{-1}(0.7) = 35.0°\text{ (1 d.p.)}\]The sine rule
ExtendedFor any triangle — not just right-angled ones — label each side with the lower-case letter of the angle opposite it:
To find a side, keep the sides on top:
\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\]To find an angle, flip it the other way up:
\[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\]Use the sine rule when you have a matching pair — a side and the angle directly opposite it — plus one more piece of information.
In triangle \(ABC\), \(b = 9.2\) cm, angle \(B = 47°\) and angle \(A = 68°\). Find side \(a\).
We have the matching pair \(b\) and \(B\), so:
\[\frac{a}{\sin 68°}=\frac{9.2}{\sin 47°}\] \[a=\frac{9.2\,\sin 68°}{\sin 47°}=11.7\text{ cm (3 s.f.)}\]The cosine rule
ExtendedUse the cosine rule when the sine rule will not work — when you do not have a matching pair. To find a side when you know two sides and the angle between them (SAS):
To find an angle when you know all three sides (SSS), rearrange to:
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]In triangle \(ABC\), \(b = 8\) cm, \(c = 5\) cm and the angle between them \(A = 60°\). Find side \(a\).
Area of a triangle
ExtendedWhen you know two sides and the angle between them, the area is:
where \(C\) is the angle enclosed by sides \(a\) and \(b\).
A triangle has two sides of \(6\) cm and \(10\) cm with an included angle of \(40°\). Find its area.
Which rule should I use?
Right-angled triangle? Use SOHCAHTOA (or Pythagoras if no angle is involved).
Got a matching pair (a side and its opposite angle)? Use the sine rule.
Two sides and the angle between them (SAS), or all three sides (SSS)? Use the cosine rule.
Two sides and the included angle, and you want the area? Use \(\frac{1}{2}ab\sin C\).
Exam-style question
\(ABC\) is a triangle. \(AB = 8.4\) cm, \(AC = 6.5\) cm and angle \(BAC = 78°\).
(a) Calculate the length of \(BC\). [3]
(b) Calculate the area of triangle \(ABC\). [2]
(c) Calculate angle \(ABC\). [3]
Show solution (a)
Two sides and the included angle → cosine rule:
\[BC^2 = 8.4^2 + 6.5^2 - 2(8.4)(6.5)\cos 78°\] \[BC^2 = 70.56 + 42.25 - 109.2\cos 78° = 90.1\] \[BC = \sqrt{90.1} = 9.49\text{ cm (3 s.f.)}\]Show solution (b)
Two sides and the included angle → area formula:
\[\text{Area}=\frac{1}{2}\times 8.4\times 6.5\times\sin 78° = 26.7\text{ cm}^2\text{ (3 s.f.)}\]Show solution (c)
Use the sine rule with the matching pair \(BC\) and angle \(A\):
\[\frac{\sin(ABC)}{6.5}=\frac{\sin 78°}{9.49}\] \[\sin(ABC)=\frac{6.5\,\sin 78°}{9.49}=0.670\] \[\angle ABC = \sin^{-1}(0.670) = 42.0°\]This is a classic Paper 4 multi-step question worth around 8 marks. It chains three tools — cosine rule, area formula and sine rule — and feeds the answer from part (a) into part (c). Keep the unrounded value of \(BC\) in your calculator for part (c); rounding to \(9.49\) too early can shift the final angle. Always write the formula and show the substitution to secure the method marks.
Common mistakes
Trigonometry at IGCSE is always in degrees. If your answers look wildly wrong, check that your calculator shows DEG at the top of the screen, not RAD or GRAD.
Opposite and adjacent depend on which angle you are using. The opposite always faces the angle; the adjacent sits between the angle and the right angle. Re-label for every new angle.
The sine rule needs a matching pair (a side and its opposite angle). If you only have two sides and the angle between them, or all three sides, you must use the cosine rule.
Carry full calculator accuracy through every step and only round the final answer. Rounding partway through a multi-step question is one of the most common ways to lose accuracy marks.
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