Trigonometry (SOHCAHTOA)

Right-angled trigonometry is one of the most powerful tools in OCR GCSE Mathematics (J560): it lets you find a missing side or angle in a right-angled triangle whenever you know two other pieces of information. The whole method is captured in the mnemonic SOHCAHTOA, which packs the three ratios — sine, cosine and tangent — into one easy-to-recall line. The exam skills are labelling the triangle correctly, choosing the right ratio, and rearranging to find the unknown, whether that is a length or an angle. Recalling and using a ratio is AO1; setting up the correct equation from a worded scenario is AO2; and applying trigonometry to bearings, ladders or heights is AO3. You also need the exact trig values for special angles.

Key Vocabulary

Term	Meaning
Hypotenuse	The longest side of a right-angled triangle, opposite the right angle
Opposite	The side directly across from the angle you are using
Adjacent	The side next to the angle (not the hypotenuse)
$\sin$sin, $\cos$cos, $\tan$tan	The three trigonometric ratios
Inverse ($\sin^{-1}$sin−1 etc.)	The function used to find an angle from a ratio
Angle of elevation	The angle measured upward from the horizontal
Angle of depression	The angle measured downward from the horizontal

Labelling the Triangle

Everything in this topic begins with labelling the three sides relative to the angle you are working with. The hypotenuse is always the longest side, lying opposite the right angle, and it never changes. The opposite is the side directly across the triangle from your chosen angle. The adjacent is the remaining side, the one next to your angle that is not the hypotenuse. The figure shows a right-angled triangle labelled from the angle $\theta$θ.

The three ratios are then $\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$sinθ=hypopp​, $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$cosθ=hypadj​ and $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$tanθ=adjopp​. The mnemonic SOH–CAH–TOA records exactly these: Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent.

The single most important decision in any trigonometry question is which of the three ratios to use, and there is a foolproof routine for making it. First identify the angle you are working from — either the angle you are given or the angle you are trying to find. Then label the three sides relative to that angle. Now look at the question and tick the two sides that are involved: the side you know and the side you want (when finding a length), or the two sides you know (when finding an angle). Finally, choose the ratio that contains exactly those two ticked sides. If the two sides are opposite and hypotenuse you use sine; if they are adjacent and hypotenuse you use cosine; if they are opposite and adjacent you use tangent. Following this routine every time removes the guesswork and is the habit that separates confident candidates from those who pick a ratio at random. It is worth stressing that the right angle itself is never the angle you work from — you always use one of the two other, non-right angles.

Finding a Missing Side

When the unknown is a side, label the triangle, choose the ratio that links the side you want with the side and angle you know, then rearrange and evaluate. The rearranging follows a simple pattern that is worth internalising. Write the ratio as a fraction equal to the trig value, then look at where the unknown sits. If the unknown side is on the top of the fraction, you find it by multiplying the known side by the trig value. If the unknown side is on the bottom of the fraction, you find it by dividing the known side by the trig value. This "top means multiply, bottom means divide" rule covers every length question without you ever needing to remember a separate rearranged formula. Throughout, keep your calculator in degrees mode — a calculator left in radians is one of the most common causes of completely wrong answers in this topic.

Worked Example 1

In a right-angled triangle the angle is $35^{\circ}$35∘ and the hypotenuse is $12$12 cm. Work out the length of the side opposite the angle, to $1$1 decimal place.

Opposite and hypotenuse means sine: $\sin 35^{\circ} = \dfrac{\text{opp}}{12}$sin35∘=12opp​. Rearranging, opp $= 12 \times \sin 35^{\circ} \approx 12 \times 0.5736 = 6.9$=12×sin35∘≈12×0.5736=6.9 cm.

Answer: $6.9$6.9 cm.

Worked Example 2

A right-angled triangle has angle $40^{\circ}$40∘ and the adjacent side is $9$9 cm. Work out the hypotenuse, to $1$1 decimal place.

Adjacent and hypotenuse means cosine: $\cos 40^{\circ} = \dfrac{9}{\text{hyp}}$cos40∘=hyp9​. Here the unknown is on the bottom, so hyp $= \dfrac{9}{\cos 40^{\circ}} \approx \dfrac{9}{0.766} = 11.7$=cos40∘9​≈0.7669​=11.7 cm.

Answer: $11.7$11.7 cm.

Common error: multiplying instead of dividing when the unknown is in the denominator. If the unknown is on the bottom, divide the known side by the ratio.

Worked Example 3

A right-angled triangle has angle $52^{\circ}$52∘ and the side adjacent to it is $7$7 cm. Work out the opposite side, to $1$1 decimal place.

Opposite and adjacent means tangent: $\tan 52^{\circ} = \dfrac{\text{opp}}{7}$tan52∘=7opp​, so opp $= 7 \times \tan 52^{\circ} \approx 7 \times 1.2799 = 9.0$=7×tan52∘≈7×1.2799=9.0 cm.

Answer: $9.0$9.0 cm.

Finding a Missing Angle

When the unknown is an angle, the process is slightly different. You still label the triangle and choose the ratio from the two sides you know, but instead of rearranging for a length you now have a known numerical ratio and need to work backwards to the angle that produced it. This is where the inverse functions come in. Each ratio has an inverse — $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 and $\tan^{-1}$tan−1 — which takes a ratio and returns the angle. On a calculator these are the second-function (or "shift") keys printed above the sine, cosine and tangent buttons. So the recipe is: write the ratio, divide out the two known sides to get a decimal, then apply the matching inverse function to that decimal. Understanding that the inverse function literally reverses the original ratio is the key idea, and it stops the common mistake of pressing the ordinary sine button when an angle is wanted.

Worked Example 4

A right-angled triangle has an opposite side of $5$5 cm and a hypotenuse of $13$13 cm. Work out the angle, to $1$1 decimal place.

Opposite and hypotenuse means sine: $\sin\theta = \dfrac{5}{13}$sinθ=135​, so $\theta = \sin^{-1}\left(\dfrac{5}{13}\right) \approx 22.6^{\circ}$θ=sin−1(135​)≈22.6∘.

Answer: $22.6^{\circ}$22.6∘.

Worked Example 5

A right-angled triangle has an adjacent side of $8$8 cm and an opposite side of $6$6 cm. Work out the angle, to $1$1 decimal place.

Opposite and adjacent means tangent: $\tan\theta = \dfrac{6}{8} = 0.75$tanθ=86​=0.75, so $\theta = \tan^{-1}(0.75) \approx 36.9^{\circ}$θ=tan−1(0.75)≈36.9∘.

Answer: $36.9^{\circ}$36.9∘.

Common error: forgetting to use the inverse function and instead typing $\sin(0.75)$sin(0.75), which gives a meaningless tiny number. To go from a ratio to an angle you must use $\sin^{-1}$sin−1, $\cos^{-1}$cos−1 or $\tan^{-1}$tan−1.

Worked Example 6

A right-angled triangle has the adjacent side $10$10 cm and the hypotenuse $14$14 cm. Work out the angle, to $1$1 decimal place.

Adjacent and hypotenuse means cosine: $\cos\theta = \dfrac{10}{14}$cosθ=1410​, so $\theta = \cos^{-1}\left(\dfrac{10}{14}\right) \approx 44.4^{\circ}$θ=cos−1(1410​)≈44.4∘.

Answer: $44.4^{\circ}$44.4∘.

Exact Trig Values

Paper 1 of J560 is a non-calculator paper, so for certain special angles you are expected to know the exact values of the trig ratios off by heart. The angles that matter are $0^{\circ}$0∘, $30^{\circ}$30∘, $45^{\circ}$45∘, $60^{\circ}$60∘ and $90^{\circ}$90∘, and the word "exact" in a question is your signal that the answer should be left as a fraction or a surd rather than as a rounded decimal. These values are not arbitrary: they come from two special right-angled triangles. Cutting an equilateral triangle in half produces a right-angled triangle containing the $30^{\circ}$30∘ and $60^{\circ}$60∘ angles, and a right-angled isosceles triangle (a square cut along its diagonal) produces the $45^{\circ}$45∘ angle. You can re-derive the values from these triangles using Pythagoras if you ever forget them, but in the exam it is far quicker to recall the table directly, so it is well worth memorising. A common pattern that aids memory is that the sine values for $0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}$0∘,30∘,45∘,60∘,90∘ can be written as $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$20​​,21​​,22​​,23​​,24​​, and the cosine values are the same list read backwards.