Trigonometric Ratios
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.
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.
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.
Key facts
- The definitions of $\sin \theta$, $\cos \theta$, and $\tan \theta$
- The relationship $\tan \theta = \frac{\sin \theta}{\cos \theta}$
- The Pythagorean identity
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
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
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.
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.
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.
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 $\cos \theta$.
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.
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 $\theta$.
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.
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).
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.
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.
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.
Work these through step-by-step. The first three focus on finding sides; the last two require the Pythagorean identity.
In a right-angled triangle, $\sin \theta = \frac{3}{5}$ and the hypotenuse is 20 cm. Find the opposite side.
Since $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$, then $\text{opposite} = \frac{3}{5} \times 20 = 12$ cm.
In a right-angled triangle, $\tan \theta = 2$ and the adjacent side is 3 cm. Find the exact opposite side.
$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{3} = 2 \Rightarrow x = 6$ cm.
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.
$\cos 15^\circ = \frac{10}{L} \Rightarrow L = \frac{10}{\cos 15^\circ} \approx 10.4$ m.
Given $\cos \theta = \frac{5}{13}$ and $\theta$ is acute, find $\sin \theta$.
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).
Given $\sin \theta = \frac{1}{\sqrt{2}}$ and $\theta$ is acute, find $\tan \theta$.
If $\sin\theta = \frac{1}{\sqrt{2}}$, then $\theta = 45^\circ$ and $\tan 45^\circ = 1$.
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 $45^\circ$.
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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.
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:
Finding the angle:
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.
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 $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
Check with identity: $\sin^2\theta + \cos^2\theta = \frac{9}{25} + \frac{16}{25} = 1$ ✓
Answer: $\theta$ is in quadrant III and $\cos\theta = \mathbf{-\frac45}$.
Notice: we didn't need to find $\theta$ itself. The quadrant analysis gives us the sign, and the ratio gives us the magnitude.
Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaJump up the platform while answering Module 2 questions. Lighter alternative to the boss.
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