What is Right Angled Trigonometry in the context of the Cambridge IGCSE Mathematics curriculum?

Right Angled Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of right-angled triangles. In the Cambridge IGCSE Mathematics curriculum, students learn to apply trigonometric ratios such as sine, cosine, and tangent to solve problems involving right-angled triangles.

How do you use trigonometric ratios to find unknown sides in a right-angled triangle?

To find unknown sides in a right-angled triangle using trigonometric ratios, identify the sides relative to the angle of interest: opposite, adjacent, and hypotenuse. Use the sine ratio (opposite/hypotenuse), cosine ratio (adjacent/hypotenuse), or tangent ratio (opposite/adjacent) depending on the given information. Solve the equation to find the unknown side.

What are the common trigonometric ratios used in Right Angled Trigonometry?

The common trigonometric ratios used in Right Angled Trigonometry are sine (sin), cosine (cos), and tangent (tan). These ratios relate the angles of a right-angled triangle to the lengths of its sides: sine is the ratio of the opposite side to the hypotenuse, cosine is the ratio of the adjacent side to the hypotenuse, and tangent is the ratio of the opposite side to the adjacent side.

How can you determine the angle of a right-angled triangle using trigonometric ratios?

To determine the angle of a right-angled triangle using trigonometric ratios, use the inverse functions of sine, cosine, or tangent. For example, if you know the lengths of the opposite and hypotenuse sides, use the inverse sine function (sin⁻¹) to find the angle. Similarly, use cos⁻¹ or tan⁻¹ for cosine and tangent ratios, respectively.

What is the importance of the Pythagorean theorem in Right Angled Trigonometry?

The Pythagorean theorem is crucial in Right Angled Trigonometry as it provides a relationship between the sides of a right-angled triangle. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is often used alongside trigonometric ratios to solve problems involving right-angled triangles in the Cambridge IGCSE Mathematics curriculum.

Why is "Calculator in radian mode" a common mistake in Right Angled Trigonometry?

Why it happens: Default mode varies by calculator. How to avoid it: Switch to DEG at the start of the paper. Spot it: 30 -0.99 in radians, 0.5 in degrees. Source: 0580/42 — recurring

Why is "Mislabelling opposite vs adjacent" a common mistake in Right Angled Trigonometry?

Why it happens: Labels depend on which angle you're working from. How to avoid it: Stand at the angle. Hypotenuse is across from the right angle. Opposite is across from the angle. Adjacent is the remaining one.

Why is "Forgetting to apply $\sin^{-1}$ when finding an angle" a common mistake in Right Angled Trigonometry?

Why it happens: Stopping at = 0.42 and writing = 0.42. How to avoid it: If you have = value, your last step is = ^-1(value).

Why is "Rounding intermediate trig values to 2 d.p." a common mistake in Right Angled Trigonometry?

Why it happens: Looks tidier. How to avoid it: Keep at least 4 d.p. (or use ans/memory) until the final answer. Round once at the end.

TrigonometryCambridge IGCSE Maths 0580 Extended

Right Angled Trigonometry

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Right-angled Trigonometry — Cambridge IGCSE 0580 Maths Extended (2026)

SOH CAH TOA. Sine, cosine and tangent ratios in right-angled triangles. Find missing sides or angles by labelling Hypotenuse, Opposite, Adjacent first.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E8.1 — Apply Pythagoras' theorem and sine, cosine and tangent ratios for acute angles to the calculation of a side or of an angle of a right-angled triangle.

SOH CAH TOA — the three ratios

Label hyp/opp/adj relative to the angle. Pick the ratio with TWO sides you have or want.

In a right-angled triangle with one acute angle $\theta$ :

$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}.$

Labelling the sides.

Hypotenuse: opposite the right angle (always the longest).
Opposite: across from $\theta$ .
Adjacent: next to $\theta$ AND the right angle.

SOH CAH TOA.

Sin = Opposite over Hypotenuse.
Cos = Adjacent over Hypotenuse.
Tan = Opposite over Adjacent.

Hypotenuse faces the right angle; opposite faces θ; adjacent sits between θ and the right angle.

Worked. Right-angled triangle, $\theta = 35\degree$ , hypotenuse $= 12\,\text{cm}$ . Find the opposite side.

We have hyp, want opp → use sine.
$\sin 35\degree = \dfrac{\text{opp}}{12} \Rightarrow \text{opp} = 12 \sin 35\degree \approx 6.88\,\text{cm}$ .

Label hyp / opp / adj first — relative to $\theta$ .
Hypotenuse = longest = opposite right angle.
Pick the ratio with the TWO sides involved.
Set calculator to DEGREE mode.

Finding a missing side

Substitute the angle and known side, then rearrange. Multiply if the unknown is in the numerator; divide if in the denominator.

Strategy.

Identify the angle and the two sides (one known, one unknown).
Pick the right ratio (sine, cosine, or tangent).
Substitute, then rearrange to isolate the unknown.

Worked 1 (unknown in numerator). Find $x$ when $\theta = 50\degree$ and adjacent $= 8\,\text{cm}$ , $x$ is opposite.

$\tan 50\degree = \dfrac{x}{8} \Rightarrow x = 8 \tan 50\degree \approx 9.53\,\text{cm}$ .

Worked 2 (unknown in denominator). Find $x$ when $\theta = 40\degree$ and opposite $= 7\,\text{cm}$ , $x$ is hypotenuse.

$\sin 40\degree = \dfrac{7}{x} \Rightarrow x = \dfrac{7}{\sin 40\degree} \approx 10.89\,\text{cm}$ .

Tip. Cross-multiply mentally: when the unknown is on top, multiply both sides by the bottom; when it's on the bottom, swap it with the trig value.

Pick the ratio matching the known and unknown sides.
Unknown on top: multiply.
Unknown on bottom: divide.
Round to 2-3 d.p. or 3 s.f.

Finding a missing angle

Use the inverse trig function: $\sin^{-1}, \cos^{-1}, \tan^{-1}$ .

When two sides are known and you need the angle, use the inverse trig function.

Worked. Right-angled triangle, opposite $= 5\,\text{cm}$ , hypotenuse $= 13\,\text{cm}$ . Find $\theta$ .

$\sin\theta = \dfrac{5}{13} \Rightarrow \theta = \sin^{-1}\left(\dfrac{5}{13}\right) \approx 22.6\degree$ .

Notation. $\sin^{-1} x$ means "the angle whose sine is $x$ " — NOT $\dfrac{1}{\sin x}$ . On most calculators, press SHIFT (or 2nd) then sin.

Common ratios to recognise.

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0\degree$	0	1	0
$30\degree$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$
$45\degree$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$
$60\degree$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$
$90\degree$	1	0	undef

Inverse trig: $\sin^{-1}, \cos^{-1}, \tan^{-1}$ .
$\sin^{-1} x \ne \dfrac{1}{\sin x}$ .
Memorise exact values for $30\degree, 45\degree, 60\degree$ .

Angles of elevation and depression

Elevation: looking UP from horizontal. Depression: looking DOWN.

Angle of elevation. From an observer's eye, the angle ABOVE the horizontal up to a target.

Angle of depression. From an observer's eye, the angle BELOW the horizontal down to a target.

Worked. From the top of a $20\,\text{m}$ tower, the angle of depression to a car on the ground is $35\degree$ . How far is the car from the tower's base?

Draw the right-angled triangle: tower (vertical) = $20$ , distance to car (horizontal) = $x$ , angle of depression at top = $35\degree$ .
The angle of depression at the top equals the angle of elevation at the car (alternate angles, parallel horizontals).
$\tan 35\degree = \dfrac{20}{x} \Rightarrow x = \dfrac{20}{\tan 35\degree} \approx 28.6\,\text{m}$ .

Tip. Always draw a clear right-angled triangle and label the angle, opposite, adjacent. Cambridge marks the diagram set-up — even rough work is worth checking.

Depression at the top = elevation at the car — they are alternate angles between parallel horizontals.

Elevation: angle UP from horizontal.
Depression: angle DOWN from horizontal.
Depression at top = elevation at base (parallel horizontals).
Always draw and label the triangle.

How it’s examined

Right-angled trig appears every Paper 2 (3-4 marks) and Paper 4 (typically as a sub-question of a longer problem worth 4-6 marks). Examiner reports flag two errors: (i) calculator in radian mode, (ii) using the wrong side label (treating the side OPPOSITE the right angle as adjacent).

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E8.1); 0580/22 May/Jun 2024 — Q9 (right-angled trig); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Right Angled Trigonometry

Step-by-step solutions to past-paper-style questions on right angled trigonometry, written exactly the way a tutor would explain them at the board.

1Find an unknown side using cosine
Core• cosine
Question
In a right-angled triangle, the hypotenuse is $10$ cm and one angle is $35\degree$ . Find the side adjacent to that angle.
Step-by-step solution
1. Step 1
  $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ .
  $\cos 35\degree = \dfrac{x}{10}$
2. Step 2
  Rearrange and compute.
  $x = 10 \cos 35\degree \approx 8.19\,\text{cm}$
Answer
$x \approx 8.19\,\text{cm}$
2Find an unknown angle
Core• Adapted from 0580/22 May/Jun 2024 Q15• inverse trig
Question
A right-angled triangle has opposite side $7$ cm and adjacent side $5$ cm. Find the angle.
Step-by-step solution
1. Step 1
  Use $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$ .
  $\tan\theta = \dfrac{7}{5}$
2. Step 2
  Apply $\tan^{-1}$ .
  $\theta = \tan^{-1}(1.4) \approx 54.5\degree$
Answer
$\theta \approx 54.5\degree$
3Combine Pythagoras and trigonometry
Extended• pythagoras
Question
A ladder of length $6$ m leans against a wall. Its foot is $2$ m from the base of the wall. Find the angle the ladder makes with the ground.
Step-by-step solution
1. Step 1
  Hypotenuse = $6$ , adjacent (along ground) = $2$ .
  $\cos\theta = \dfrac{2}{6} = \dfrac{1}{3}$
2. Step 2
  Compute.
  $\theta = \cos^{-1}\left(\dfrac{1}{3}\right) \approx 70.5\degree$
Answer
$\theta \approx 70.5\degree$
4Angle of elevation
Extended• elevation
Question
From a point $50$ m from the base of a tower, the angle of elevation of the top is $32\degree$ . Find the height of the tower.
Step-by-step solution
1. Step 1
  Horizontal = adjacent, height = opposite. Use $\tan$ .
  $\tan 32\degree = \dfrac{h}{50}$
2. Step 2
  Solve for $h$ .
  $h = 50\tan 32\degree \approx 31.2\,\text{m}$
Answer
$h \approx 31.2\,\text{m}$
5Find an unknown side using sine
Core• Adapted from 0580/22 Feb/Mar 2023 Q9• sine, find side
Question
In a right-angled triangle, the hypotenuse is $14$ cm and one angle is $42\degree$ . Find the side opposite to that angle.
Step-by-step solution
1. Step 1
  Opposite and hypotenuse are involved, so use $\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$ .
  $\sin 42\degree = \dfrac{x}{14}$
2. Step 2
  Rearrange.
  $x = 14\sin 42\degree \approx 9.37\,\text{cm}$
Answer
$x \approx 9.37$ cm
Examiner tip
Always identify the labelled angle first, then label opposite/adjacent/hypotenuse relative to it. Picking the wrong ratio is the most common reason for losing the method mark here.
6Angle of depression from a cliff
Extended• Adapted from 0580/42 May/Jun 2023 Q11• depression
Question
From the top of a cliff $85$ m high, the angle of depression to a boat at sea is $24\degree$ . Find the horizontal distance from the foot of the cliff to the boat.
Step-by-step solution
1. Step 1
  The angle of depression equals the angle of elevation from the boat to the top of the cliff (alternate angles). The cliff height is opposite, the horizontal distance is adjacent.
  $\tan 24\degree = \dfrac{85}{d}$
2. Step 2
  Rearrange for $d$ .
  $d = \dfrac{85}{\tan 24\degree} \approx 190.9\,\text{m}$
Answer
$d \approx 191$ m (3 s.f.)
Examiner tip
The 2023 examiner report notes many candidates put the $85$ on the adjacent side. Always sketch the right-angled triangle, marking the depression angle at the top.
7Find an angle using inverse sine
Core• inverse trig, find angle
Question
A right-angled triangle has hypotenuse $20$ cm and opposite side $13$ cm relative to the angle $\theta$ . Find $\theta$ .
Step-by-step solution
1. Step 1
  Opposite and hypotenuse are known, so use $\sin$ .
  $\sin\theta = \dfrac{13}{20} = 0.65$
2. Step 2
  Apply $\sin^{-1}$ .
  $\theta = \sin^{-1}(0.65) \approx 40.5\degree$
Answer
$\theta \approx 40.5\degree$
Examiner tip
Stopping at $\sin\theta = 0.65$ and writing $\theta = 0.65$ is the most common error. The final step is always $\theta = \sin^{-1}(\text{ratio})$ .
8Find a side, then use it in another triangle
Extended• Adapted from 0580/42 Oct/Nov 2023 Q14• multi-step
Question
Triangle $ABC$ has a right angle at $B$ , $AB = 8$ cm and $\angle BAC = 35\degree$ . Point $D$ lies on $BC$ such that $\angle BAD = 20\degree$ . Find the length of $DC$ .
Step-by-step solution
1. Step 1
  In triangle $ABC$ , find $BC$ using $\tan 35\degree = \dfrac{BC}{AB}$ .
  $BC = 8\tan 35\degree \approx 5.602\,\text{cm}$
2. Step 2
  In triangle $ABD$ , $\tan 20\degree = \dfrac{BD}{AB}$ .
  $BD = 8\tan 20\degree \approx 2.912\,\text{cm}$
3. Step 3
  Subtract to find $DC$ .
  $DC = BC - BD \approx 5.602 - 2.912 \approx 2.69\,\text{cm}$
Answer
$DC \approx 2.69$ cm (3 s.f.)
Examiner tip
Keep at least four decimal places when reusing intermediate values — the 2023 mark scheme penalises premature rounding that changes the final answer beyond the accuracy tolerance.
9A* multi-step: flagpole and shadow
Challenge• Adapted from 0580/42 May/Jun 2024 Q21• multi-step, practical
Question
A vertical flagpole stands on horizontal ground. When the sun's angle of elevation is $58\degree$ , the shadow on the ground is $9.5$ m long. Later, when the sun has moved, the shadow becomes $14.2$ m long. Find (a) the height of the flagpole, and (b) the sun's new angle of elevation.
Step-by-step solution
1. Step 1
  (a) Height $h$ is opposite the elevation angle, shadow is adjacent.
  $\tan 58\degree = \dfrac{h}{9.5}$
2. Step 2
  Solve.
  $h = 9.5 \tan 58\degree \approx 15.20\,\text{m}$
3. Step 3
  (b) Use the same height with the new shadow to find the new angle.
  $\tan\alpha = \dfrac{15.20}{14.2} \approx 1.0705$
4. Step 4
  Apply $\tan^{-1}$ .
  $\alpha = \tan^{-1}(1.0705) \approx 46.9\degree$
Answer
(a) $h \approx 15.2$ m (b) $\alpha \approx 46.9\degree$
Examiner tip
Carry the unrounded value of $h$ from (a) into (b). The 2024 examiner report notes that candidates who rounded $h$ to $15.2$ in (b) often dropped the final accuracy mark.
10Find an unknown side using tangent
Core• tangent, find side
Question
A right-angled triangle has an angle of $28\degree$ with the adjacent side $12$ cm. Find the length of the opposite side.
Step-by-step solution
1. Step 1
  Both legs are involved (no hypotenuse), so use $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$ .
  $\tan 28\degree = \dfrac{x}{12}$
2. Step 2
  Rearrange.
  $x = 12 \tan 28\degree \approx 6.38\,\text{cm}$
Answer
$x \approx 6.38$ cm
Examiner tip
Pick the trig ratio that involves both the known side and the unknown side. With opposite and adjacent, $\tan$ is the only option.

Key Formulae — Right Angled Trigonometry

The formulae you need to memorise for right angled trigonometry on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Sine ratio
$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
When to use
Opposite or hypotenuse known/required.
Cosine ratio
$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
When to use
Adjacent or hypotenuse known/required.
Tangent ratio
$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$
When to use
When both legs are involved (no hypotenuse).
Inverse trigonometric functions
$\theta = \sin^{-1}(x),\ \cos^{-1}(x),\ \tan^{-1}(x)$
When to use
Use to recover angle from a known ratio.

Key Definitions and Keywords — Right Angled Trigonometry

Definitions to memorise and the exact keywords mark schemes credit for right angled trigonometry answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Hypotenuse
Examiner keyword
The longest side of a right-angled triangle, opposite the right angle.
Opposite
Examiner keyword
Side directly across from the angle in question.
Adjacent
Examiner keyword
The leg next to the angle in question (not the hypotenuse).
Angle of elevation
Examiner keyword
The angle measured from the horizontal upward to a higher point.
Angle of depression
Examiner keyword
The angle measured from the horizontal downward to a lower point.

Common Mistakes and Misconceptions — Right Angled Trigonometry

The traps other students keep falling into on right angled trigonometry questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Calculator in radian mode
0580/42 — recurring
Why it happens
Default mode varies by calculator.
How to avoid it
Switch to DEG at the start of the paper. Spot it: $\sin 30 \approx -0.99$ in radians, $0.5$ in degrees.
✕Mislabelling opposite vs adjacent
Why it happens
Labels depend on which angle you're working from.
How to avoid it
Stand at the angle. Hypotenuse is across from the right angle. Opposite is across from the angle. Adjacent is the remaining one.
✕Forgetting to apply $\sin^{-1}$ when finding an angle
Why it happens
Stopping at $\sin\theta = 0.42$ and writing $\theta = 0.42$ .
How to avoid it
If you have $\sin\theta =$ value, your last step is $\theta = \sin^{-1}(\text{value})$ .
✕Rounding intermediate trig values to 2 d.p.
Why it happens
Looks tidier.
How to avoid it
Keep at least 4 d.p. (or use $\text{ans}$ /memory) until the final answer. Round once at the end.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Right Angled Trigonometry — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Right Angled Trigonometry

Short Study Notes in the form of Slides

Short Study Notes — Right Angled Trigonometry

Detailed Notes

What you’ll learn

How it’s examined

1Find an unknown side using cosine

2Find an unknown angle

3Combine Pythagoras and trigonometry

4Angle of elevation

5Find an unknown side using sine

6Angle of depression from a cliff

7Find an angle using inverse sine

8Find a side, then use it in another triangle

9A* multi-step: flagpole and shadow

10Find an unknown side using tangent

Sine ratio

Cosine ratio

Tangent ratio

Inverse trigonometric functions

Hypotenuse

Opposite

Adjacent

Angle of elevation

Angle of depression

✕Calculator in radian mode

✕Mislabelling opposite vs adjacent

✕Forgetting to apply sin⁡−1\sin^{-1}sin−1 when finding an angle

✕Rounding intermediate trig values to 2 d.p.

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Right Angled Trigonometry — frequently asked questions

✕Forgetting to apply $\sin^{-1}$ when finding an angle