Engineers designing bridges, pilots navigating flight paths, and architects planning roofs all rely on the same three ratios: sine, cosine, and tangent. In this lesson you will learn how these ratios are defined from both right-angled triangles and the unit circle, and how to use them to solve real-world problems.
Today's hook, A ladder leans against a wall, making an angle of 60° with the ground. The ladder is 4 metres long. How high up the wall does the ladder reach? Try to solve this without a calculator.
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
A ladder leans against a wall, making an angle of $60^\circ$ with the ground. The ladder is 4 metres long. Without using a calculator, how high up the wall does the ladder reach? Use what you know about special triangles.
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The three ratios
Work through the core explanation before applying it.
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The three ratios
+5 XP to read
There are only three trigonometric ratios in a right-angled triangle. Every bridge, roof, and flight path calculation reduces to one of these. They are properties of the angle, not the triangle's size.
In any right-angled triangle, sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. These ratios stay constant when the triangle is scaled.
SOH · CAH · TOA
Sine
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$. Use when you know the hypotenuse and want the opposite side.
Cosine
$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. Use when you know the hypotenuse and want the adjacent side.
Tangent
$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin \theta}{\cos \theta}$. Use when you know both legs and want the angle.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The definitions of $\sin \theta$, $\cos \theta$, and $\tan \theta$
The relationship $\tan \theta = \frac{\sin \theta}{\cos \theta}$
The Pythagorean identity
Understand
Concepts
How trig ratios connect angles to side lengths in right triangles
How the unit circle and triangle definitions are consistent
Why the Pythagorean identity holds for all angles
Can do
Skills
Find missing sides and angles in right-angled triangles
Use the Pythagorean identity to find missing trig ratios
Solve applied problems using sine, cosine, and tangent
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Key terms
Sine$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$, the ratio of the opposite side to the hypotenuse.
Cosine$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, the ratio of the adjacent side to the hypotenuse.
Tangent$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$, the ratio of opposite to adjacent.
HypotenuseThe longest side of a right-angled triangle, opposite the right angle.
Pythagorean identity$\sin^2 \theta + \cos^2 \theta = 1$ for every angle $\theta$.
Roof pitchThe steepness of a roof, measured as the tangent of the roof angle.
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How trigonometric ratios connect angles to sides
Work through the core explanation before applying it.
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How trigonometric ratios connect angles to sides
core concept
Trigonometric ratios are functions of the angle, not the triangle's size. If you scale a right-angled triangle up or down, the ratios stay the same because every side scales by the same factor.
From a point on the unit circle, dropping a perpendicular to the $x$-axis creates a right triangle whose hypotenuse is the radius 1. The $x$-coordinate is $\cos\theta$ and the $y$-coordinate is $\sin\theta$. This is why the triangle definition and the unit circle definition give exactly the same values for acute angles.
The Pythagorean identity follows directly from the unit circle:
This identity is a free extra equation: if you know one of $\sin\theta$ or $\cos\theta$, you can find the other without ever solving a triangle.
Why trig ratios don't care about triangle size. The angle is the same whether you draw a tiny triangle or a giant one. Scaling every side by the same factor $k$ leaves all three ratios unchanged, numerator and denominator both multiply by $k$, so $k$ cancels. This is why a ratio measured on a model is exactly valid for a real bridge or aircraft.
SOH: $\sin\theta = \frac{\text{opp}}{\text{hyp}}$ · CAH: $\cos\theta = \frac{\text{adj}}{\text{hyp}}$ · TOA: $\tan\theta = \frac{\text{opp}}{\text{adj}}$; $\tan\theta = \frac{\sin\theta}{\cos\theta}$, tangent can always be derived from sine and cosine
Pause, copy SOH CAH TOA and the derived identity $\tan\theta = \sin\theta/\cos\theta$ into your book.
Quick check: In a right triangle, which ratio uses the opposite side and the hypotenuse?
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Worked Example, Finding a missing side
+5 XP for trying first
We just saw that SOH CAH TOA connects the known angle to two sides, and $\tan\theta = \sin\theta/\cos\theta$. That raises a question: when a side is unknown, which ratio do I choose and how do I set up the equation? This card answers it → identify which two sides are involved (opp/hyp → SOH), write the equation, then isolate the unknown side.
A right-angled triangle has an angle of $30^\circ$, a hypotenuse of 8 cm, and an unknown side opposite the $30^\circ$ angle. Find the missing side.
Your turn first. Try it yourself before viewing the solution.
SOLUTION
3 marks · HSC band 4+
Given:
Hypotenuse = 8 cm
Angle opposite unknown side = $30^\circ$
Choose the correct ratio. We know the hypotenuse and want the opposite side, so use sine:
Solve. From exact values, $\sin 30^\circ = \frac12$:
$$\frac12 = \frac{x}{8} \;\Rightarrow\; x = 4$$
Answer: The missing side is 4 cm.
Step 1: identify which two sides are involved (here: opposite and hypotenuse → use sine); Step 2: write the ratio equation, substitute known values
Pause, copy the two-step procedure for finding a missing side: (1) identify which two sides are involved to choose the ratio, (2) write the ratio equation and substitute into your book.
We just saw how to find a missing side using SOH CAH TOA with a known angle. That raises a question: what if you know one ratio (like $\tan\theta$) but need the other two, and you don't have a right triangle drawn? This card answers it → draw a reference triangle using the known ratio's numerator and denominator, then use Pythagoras to find the hypotenuse.
Given $\tan \theta = \frac34$ and $\theta$ is acute, find the exact value of $\cos \theta$.
Your turn first. Try it yourself before viewing the solution.
SOLUTION
3 marks · HSC band 5+
Method 1: Right-triangle approach.
Draw a right triangle with opposite side 3 and adjacent side 4. By Pythagoras:
Since $\tan \theta = \frac34$, we have $\sin \theta = \frac35$ and $\cos \theta = \frac45$ from the same triangle. Alternatively, use $\sin^2\theta + \cos^2\theta = 1$ with $\sin\theta = \frac35$.
Answer: $\cos \theta = \mathbf{\frac45}$.
When you know $\tan\theta$, draw a right triangle with opp and adj, use Pythagoras to find the hypotenuse; From the 3-4-5 right triangle: $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$, $\tan\theta = \frac{3}{4}$
Pause, copy the reference-triangle method: when $\tan\theta = \text{opp}/\text{adj}$, label the triangle, use Pythagoras for the hypotenuse, then read off all three ratios into your book.
True or false: $\sin^2\theta + \cos^2\theta = 1$ is true only for acute angles (0° to 90°).
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Worked Example, Finding an angle
+5 XP for trying first
We just saw how to extract all three ratios from a known ratio using a reference triangle. That raises a question: what if the sides are known but the angle is unknown, how do we reverse the ratio to find the angle? This card answers it → apply the inverse trig function ($\cos^{-1}$, $\sin^{-1}$, or $\tan^{-1}$) after simplifying the ratio.
In a right-angled triangle, the adjacent side is 6 cm and the hypotenuse is 12 cm. Find the angle $\theta$.
Your turn first. Try it yourself before viewing the solution.
SOLUTION
2 marks · HSC band 3+
Choose the correct ratio. We know adjacent and hypotenuse, so use cosine:
To find an angle: set up the ratio, simplify the fraction, apply inverse trig ($\cos^{-1}$, $\sin^{-1}$, $\tan^{-1}$); Exact value: $\cos^{-1}\!\left(\frac{1}{2}\right) = 60^\circ = \frac{\pi}{3}$ rad
Pause, copy the angle-finding procedure (set up ratio → simplify → apply inverse trig) and the exact value $\cos^{-1}(1/2) = 60° = \pi/3$ into your book.
Which one is the odd one out, it does NOT belong with the others?
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Common traps
Meet the mistakes that cost marks, then do it yourself.
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Common traps
We just saw that inverse trig gives the principal value for an angle. That raises a question: what are the most common errors students make on these questions in exams? This card answers it → the two traps: labelling sides relative to the wrong angle, and forgetting that $\cos^{-1}$ and $\sin^{-1}$ give principal values only (check the quadrant for obtuse angles).
Trap 1, Wrong ratio for the given sides
Students often pick sine when they should use cosine, or vice versa. Check carefully which two sides you know and which you need. Ask: do I know hypotenuse + another side? If yes, use sine or cosine. If not, use tangent.
Remember: SOH · CAH · TOA, label the sides first.
Trap 2, Principal value only
When using $\cos^{-1}x$ or $\sin^{-1}x$ on a calculator, the answer is always in the principal range. If the triangle is in a different quadrant, you may need $180^\circ - \theta$ or $360^\circ - \theta$. For right triangles this isn't an issue, but it becomes critical once you work with obtuse and reflex angles.
Trap 3, Confusing $\sin^2\theta$ with $\sin(\theta^2)$
Always remember that $\sin^2\theta = (\sin\theta)^2$. This is not $\sin(\theta^2)$. The notation $\sin^2\theta$ is shorthand for the square of the sine of $\theta$.
Always label sides (opposite, adjacent, hypotenuse) relative to your angle before choosing a ratio; $\cos^{-1}$ and $\sin^{-1}$ give principal values, check the quadrant if the angle could be obtuse
Pause, copy both traps: (1) always label opp/adj/hyp relative to your angle before choosing a ratio; (2) $\cos^{-1}$ and $\sin^{-1}$ give principal values only, check the quadrant if the angle may be obtuse into your book.
In your own words (10 words max): what does SOH·CAH·TOA help you remember?
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Drill, build fluency
Work through the core explanation before applying it.
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Drill, build fluency
+5 XP for 5 correct
Work these through step-by-step. The first three focus on finding sides; the last two require the Pythagorean identity.
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In a right-angled triangle, $\sin \theta = \frac{3}{5}$ and the hypotenuse is 20 cm. Find the opposite side.
Show answer
12 cm Since $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$, then $\text{opposite} = \frac{3}{5} \times 20 = 12$ cm.
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In a right-angled triangle, $\tan \theta = 2$ and the adjacent side is 3 cm. Find the exact opposite side.
Show answer
6 cm $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{3} = 2 \Rightarrow x = 6$ cm.
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A ramp makes a $15^\circ$ angle with the ground. The horizontal distance is 10 m. Find the ramp length (hypotenuse) to 1 decimal place.
Show answer
10.4 m $\cos 15^\circ = \frac{10}{L} \Rightarrow L = \frac{10}{\cos 15^\circ} \approx 10.4$ m.
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Given $\cos \theta = \frac{5}{13}$ and $\theta$ is acute, find $\sin \theta$.
Show answer
$\frac{12}{13}$ Using $\sin^2\theta + \cos^2\theta = 1$: $\sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\sin\theta = \frac{12}{13}$ (positive since acute).
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Given $\sin \theta = \frac{1}{\sqrt{2}}$ and $\theta$ is acute, find $\tan \theta$.
Show answer
1 If $\sin\theta = \frac{1}{\sqrt{2}}$, then $\theta = 45^\circ$ and $\tan 45^\circ = 1$.
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Revisit, the ladder problem
Run the quick drill and copy the summary into your book.
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Revisit, the ladder problem
+5 XP for checking
Return to your original answer from Section 01. A 4-metre ladder makes a $60^\circ$ angle with the ground. Using $\sin 60^\circ = \frac{\sqrt{3}}{2}$, the height is:
To 1 decimal place: $2\sqrt{3} \approx 3.5$ metres.
How does your gut answer compare? Did you think the height would be 2 m (half the ladder), 3 m, or something else? The actual answer is closer to 3.5 m, notice how the height is actually more than half the ladder length because $60^\circ$ is steeper than $45^\circ$.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 4–5
Two possible cosine values
Given $\sin \theta = \frac{3}{5}$, find the two possible values of $\cos \theta$.
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Working:
Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:
Answer: $\cos\theta = \mathbf{\frac45}$ or $\mathbf{-\frac45}$.
Why two values? If $\sin\theta = \frac35$ (positive), $\theta$ could be in quadrant I ($\cos\theta > 0$) or quadrant II ($\cos\theta < 0$).
In HSC exams, always consider both quadrants unless the angle is restricted.
ApplyBand 5–6
Ramp problem
A wheelchair ramp must have a gradient no steeper than $1:12$ (rise : run). If the ramp makes an angle $\theta$ with the ground, find $\tan\theta$ and hence $\theta$ to the nearest degree.
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Working:
A gradient of $1:12$ means $\frac{\text{rise}}{\text{run}} = \frac{1}{12}$. By definition:
Answer: $\tan\theta = \mathbf{\frac{1}{12}}$ and $\theta \approx \mathbf{5^\circ}$ (nearest degree).
Note: this is a shallow angle. Many real ramps are around 4–5°, which is why they need to be long.
AnalyseBand 6
Find cosine from sine and tangent
Given $\sin \theta = -\frac{3}{5}$ and $\tan \theta = \frac{3}{4}$, determine the exact value of $\cos \theta$. State which quadrant $\theta$ lies in, with reasoning.
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Step 1: Identify the quadrant.
$\sin\theta = -\frac35$ → $\theta$ is in QIII or QIV (sine negative)
$\tan\theta = \frac34 > 0$ → $\theta$ is in QI or QIII (tangent positive)
The only quadrant where both conditions hold is quadrant III.
Step 2: Find $\cos\theta$.
Using $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
