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This lesson introduces right-angled triangle trigonometry for AQA GCSE Mathematics. You will learn how to label the sides of a right-angled triangle relative to a given angle, use the sine, cosine and tangent ratios to find missing sides and angles, and apply trigonometry in real-world problems. Trigonometry is a core Higher-tier topic but also appears on Foundation tier.
Before using trigonometry, you must label the three sides of a right-angled triangle relative to the angle you are working with (call it angle theta).
| Side | Abbreviation | Position |
|---|---|---|
| Hypotenuse | Hyp | The longest side, opposite the right angle |
| Opposite | Opp | The side opposite the angle theta |
| Adjacent | Adj | The side next to the angle theta (not the hypotenuse) |
graph TD
A[Choose angle theta] --> B[Hypotenuse: opposite the right angle]
A --> C[Opposite: opposite angle theta]
A --> D[Adjacent: next to angle theta, not hypotenuse]
Exam Tip: The hypotenuse never changes — it is always opposite the right angle. But the "opposite" and "adjacent" sides depend on which angle you are working with. Always label relative to the angle in the question.
SOHCAHTOA is the mnemonic for remembering the three ratios:
| Ratio | Formula | Mnemonic |
|---|---|---|
| Sine | sin(theta) = Opposite / Hypotenuse | Some Old Horses |
| Cosine | cos(theta) = Adjacent / Hypotenuse | Can Always Hear |
| Tangent | tan(theta) = Opposite / Adjacent | Their Owners Approach |
| Sides Involved | Use |
|---|---|
| Opp and Hyp | Sin |
| Adj and Hyp | Cos |
| Opp and Adj | Tan |
In a right-angled triangle, the angle is 35 degrees and the hypotenuse is 12 cm. Find the opposite side.
Solution: sin(35) = Opp / 12 Opp = 12 x sin(35) Opp = 12 x 0.5736... Opp = 6.88 cm (2 d.p.)
In a right-angled triangle, the angle is 50 degrees and the hypotenuse is 20 cm. Find the adjacent side.
Solution: cos(50) = Adj / 20 Adj = 20 x cos(50) Adj = 20 x 0.6428... Adj = 12.86 cm (2 d.p.)
In a right-angled triangle, the angle is 62 degrees and the adjacent side is 8 cm. Find the opposite side.
Solution: tan(62) = Opp / 8 Opp = 8 x tan(62) Opp = 8 x 1.8807... Opp = 15.05 cm (2 d.p.)
In a right-angled triangle, the angle is 28 degrees and the opposite side is 7 cm. Find the hypotenuse.
Solution: sin(28) = 7 / Hyp Hyp = 7 / sin(28) Hyp = 7 / 0.4695... Hyp = 14.91 cm (2 d.p.)
Exam Tip: When the unknown side is on the bottom of the fraction (the denominator), you need to rearrange. If sin(theta) = Opp / Hyp, then Hyp = Opp / sin(theta). Many students make errors here — practise the rearrangement.
To find a missing angle, use the inverse trigonometric functions: sin to the power of -1, cos to the power of -1, or tan to the power of -1 (written as arcsin, arccos, arctan on some calculators).
In a right-angled triangle, the opposite side is 6 cm and the hypotenuse is 10 cm. Find the angle.
Solution: sin(theta) = 6 / 10 = 0.6 theta = inverse sin(0.6) theta = 36.9 degrees (1 d.p.)
In a right-angled triangle, the adjacent side is 5 cm and the opposite side is 9 cm. Find the angle.
Solution: tan(theta) = 9 / 5 = 1.8 theta = inverse tan(1.8) theta = 60.9 degrees (1 d.p.)
In a right-angled triangle, the adjacent side is 8 cm and the hypotenuse is 11 cm. Find the angle.
Solution: cos(theta) = 8 / 11 = 0.7272... theta = inverse cos(0.7272...) theta = 43.3 degrees (1 d.p.)
Exam Tip: Make sure your calculator is in degree mode, not radian mode. If you get an unexpected answer (especially a very small decimal), check the mode setting. On most scientific calculators, look for "D" or "DEG" on the display.
A person stands 50 m from the base of a building and looks up at the top. The angle of elevation is 32 degrees. Find the height of the building.
Solution: tan(32) = height / 50 height = 50 x tan(32) height = 50 x 0.6249... height = 31.2 m (1 d.p.)
A ship sails 15 km on a bearing of 060 degrees. How far east has it travelled?
Solution: The eastward distance is the side opposite to the angle measured from north. In the right-angled triangle formed: sin(60) = east / 15 east = 15 x sin(60) = 15 x 0.8660... = 13.0 km (1 d.p.)
Exam Tip: In bearing questions, always draw a diagram with a north line. The angle in the triangle is often not the bearing itself but needs to be calculated from it. Take care to identify which angle you are using.
You should memorise these exact values:
| Angle | sin | cos | tan |
|---|---|---|---|
| 0 degrees | 0 | 1 | 0 |
| 30 degrees | 1/2 | (root 3)/2 | 1/(root 3) or (root 3)/3 |
| 45 degrees | (root 2)/2 | (root 2)/2 | 1 |
| 60 degrees | (root 3)/2 | 1/2 | root 3 |
| 90 degrees | 1 | 0 | undefined |
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