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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A ladder leans against a wall, making an angle of $60^\circ$ 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, using what you know about special triangles.
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Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
In a right-angled triangle, the three basic trigonometric ratios are defined as follows:
| Ratio | Definition | Memory aid |
|---|---|---|
| $\sin \theta$ | $\frac{\text{opposite}}{\text{hypotenuse}}$ | SOH |
| $\cos \theta$ | $\frac{\text{adjacent}}{\text{hypotenuse}}$ | CAH |
| $\tan \theta$ | $\frac{\text{opposite}}{\text{adjacent}}$ | TOA |
These ratios are properties of the angle, not the size of the triangle. If you scale a triangle up or down, the ratios stay the same because all sides scale proportionally.
If you place a right-angled triangle inside the unit circle with the hypotenuse as the radius, the definitions match perfectly:
This is why the unit circle definitions and the triangle definitions never contradict each other — the unit circle is just a triangle with hypotenuse $1$.
From the unit circle equation $x^2 + y^2 = 1$, substituting $x = \cos \theta$ and $y = \sin \theta$ gives:
$$\sin^2 \theta + \cos^2 \theta = 1$$
This identity is true for every angle $\theta$. It allows you to find one trig ratio if you know the other, provided you also know which quadrant the angle is in.
When you take the square root, remember to include $\pm$, then use ASTC to choose the correct sign.
🧮 Worked Examples
🧪 Activities
1 Hypotenuse = 10, angle = $30^\circ$. Find the opposite side.
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2 Adjacent = 5, angle = $45^\circ$. Find the opposite side.
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3 Opposite = 12, adjacent = 5. Find the angle (to nearest degree).
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4 Hypotenuse = 13, opposite = 5. Find $\cos \theta$.
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1 $\sin \theta = \frac{5}{13}$, $\theta$ acute. Find $\cos \theta$.
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2 $\cos \theta = -\frac{8}{17}$, $\theta$ in QII. Find $\sin \theta$.
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3 $\tan \theta = \frac{8}{15}$, $\theta$ acute. Find $\sin \theta$ and $\cos \theta$.
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Earlier you were asked: A ladder 4 m long leans against a wall at $60^\circ$ to the ground. How high up the wall does it reach?
We need the side opposite the $60^\circ$ angle, and we know the hypotenuse is 4 m. Using sine:
$$\sin 60^\circ = \frac{\text{height}}{4}$$
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
$$\text{height} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.46 \text{ m}$$
This is a classic application of exact values from special triangles. No calculator needed — just SOHCAHTOA and the exact value of $\sin 60^\circ$.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. A right-angled triangle has hypotenuse 17 cm and one side 8 cm. Find the two possible values of $\cos \theta$ if $\theta$ is one of the acute angles. Explain why both are positive. 3 MARKS
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9. A ramp is designed so that for every 1 metre of vertical rise, there are 12 metres of horizontal run. (a) Find the angle of inclination of the ramp to the nearest degree. (b) If the ramp must be 5 m long along the slope, how high does it rise? 4 MARKS
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10. If $\sin \theta = \frac{1}{\sqrt{5}}$ and $\tan \theta < 0$, find the exact value of $\cos \theta$. Justify your choice of sign. 3 MARKS
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Answer in your workbook.
1. Opposite = $10 \times \sin 30^\circ = 5$
2. Opposite = $5 \times \tan 45^\circ = 5$
3. $\tan \theta = \frac{12}{5} \Rightarrow \theta \approx 67^\circ$
4. Adjacent = $\sqrt{169 - 25} = 12$, so $\cos \theta = \frac{12}{13}$
1. $\cos \theta = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$ (positive because acute)
2. $\sin \theta = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$ (positive in QII)
3. Hypotenuse = $\sqrt{64 + 225} = 17$. $\sin \theta = \frac{8}{17}$, $\cos \theta = \frac{15}{17}$.
1. A — SOH: opposite / hypotenuse.
2. A — 3-4-5 triangle, $\sin = \frac{3}{5}$.
3. A — Definition of tangent.
4. A — Adjacent = 12, so $\cos = \frac{12}{13}$.
5. A — Pythagorean identity equals 1 for all angles.
Q8 (3 marks): Missing side = $\sqrt{289 - 64} = 15$ [1]. If $\theta$ is adjacent to 8, $\cos \theta = \frac{8}{17}$ [0.5]. If $\theta$ is adjacent to 15, $\cos \theta = \frac{15}{17}$ [0.5]. Both are positive because acute angles always have positive cosine [1].
Q9 (4 marks): (a) $\tan \theta = \frac{1}{12}$, so $\theta = \tan^{-1}\left(\frac{1}{12}\right) \approx 5^\circ$ [2]. (b) $\sin \theta = \frac{\text{rise}}{5}$, so rise $= 5 \times \sin 5^\circ \approx 0.44$ m [2].
Q10 (3 marks): $\cos^2 \theta = 1 - \frac{1}{5} = \frac{4}{5}$, so $\cos \theta = \pm \frac{2}{\sqrt{5}}$ [1]. Since $\sin \theta > 0$ and $\tan \theta < 0$, $\cos \theta$ must be negative (QII) [1]. Therefore $\cos \theta = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$ [1].
Trigonometric Ratios
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