The Fundamental Idea#
In a right-angled triangle, the angle determines the ratio between any two sides. No matter how big or small the triangle is, if the angle is the same, the ratios are the same. This is what trigonometry measures.
1. The Three Ratios (SOH CAH TOA)#
Stand at the angle $\theta$. From your perspective, label the sides:
- Opposite: The side directly across from you
- Adjacent: The side next to you (not the hypotenuse)
- Hypotenuse: The longest side (always opposite the $90°$)
| Ratio | Formula | Mnemonic |
|---|---|---|
| $\sin\theta$ | $\frac{\text{Opposite}}{\text{Hypotenuse}}$ | SOH |
| $\cos\theta$ | $\frac{\text{Adjacent}}{\text{Hypotenuse}}$ | CAH |
| $\tan\theta$ | $\frac{\text{Opposite}}{\text{Adjacent}}$ | TOA |
Key relationship: $\tan\theta = \frac{\sin\theta}{\cos\theta}$ — because $\frac{O/H}{A/H} = \frac{O}{A}$.
2. The Special Angles#
You MUST know these values without a calculator:
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $0°$ | $0$ | $1$ | $0$ |
| $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
| $45°$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
| $90°$ | $1$ | $0$ | undefined |
Where do these come from?#
The 45° triangle: An isosceles right triangle with legs = 1. Hypotenuse = $\sqrt{2}$.
The 30°-60° triangle: Half an equilateral triangle with side = 2. Short side = 1, long side = $\sqrt{3}$.
3. Reciprocal Ratios#
| Ratio | Definition |
|---|---|
| $\text{cosec}\,\theta = \frac{1}{\sin\theta}$ | $\frac{\text{Hyp}}{\text{Opp}}$ |
| $\sec\theta = \frac{1}{\cos\theta}$ | $\frac{\text{Hyp}}{\text{Adj}}$ |
| $\cot\theta = \frac{1}{\tan\theta}$ | $\frac{\text{Adj}}{\text{Opp}}$ |
These are just the “flipped” versions of sin, cos, and tan.
4. Solving for a Side#
The Method#
- Label the sides (Opp, Adj, Hyp) from the given angle.
- Choose the ratio that connects the side you HAVE with the side you WANT.
- Set up the equation and solve.
Worked Example 1: Finding the opposite#
In a right triangle with hypotenuse = 10 and angle = $35°$, find the opposite side.
$\sin 35° = \frac{\text{Opp}}{10}$
$\text{Opp} = 10 \sin 35° = 10(0.5736) = 5.74$
Worked Example 2: Finding the adjacent#
In a right triangle with opposite = 7 and angle = $50°$, find the adjacent side.
$\tan 50° = \frac{7}{\text{Adj}}$
$\text{Adj} = \frac{7}{\tan 50°} = \frac{7}{1.1918} = 5.87$
Worked Example 3: Finding the hypotenuse#
In a right triangle with adjacent = 12 and angle = $40°$, find the hypotenuse.
$\cos 40° = \frac{12}{\text{Hyp}}$
$\text{Hyp} = \frac{12}{\cos 40°} = \frac{12}{0.7660} = 15.67$
5. Solving for an Angle#
When you know two sides, use the inverse function to find the angle.
Worked Example#
In a right triangle with opposite = 5 and hypotenuse = 13, find $\theta$.
$\sin\theta = \frac{5}{13} = 0.3846$
$\theta = \sin^{-1}(0.3846) = 22.6°$
Calculator key: Press
SHIFTthenSIN(or2ndthenSIN) to get $\sin^{-1}$.
6. The Pythagoras Connection#
Trigonometry and Pythagoras work together. If you have two sides, you can always find the third using:
$$ a^2 + b^2 = c^2 $$When to use which?#
| You have | You want | Use |
|---|---|---|
| Two sides | Third side | Pythagoras |
| One side + one angle | Another side | Trig ratio |
| Two sides | An angle | Inverse trig |
Worked Example: Combined#
A ladder of length 5m leans against a wall. The foot is 3m from the wall. Find the angle with the ground and the height it reaches.
Height (Pythagoras): $h^2 + 3^2 = 5^2 \Rightarrow h = \sqrt{25 - 9} = 4$ m
Angle: $\cos\theta = \frac{3}{5} \Rightarrow \theta = \cos^{-1}(0.6) = 53.1°$
7. Problems of Elevation and Depression#
- Angle of elevation: Looking UP from horizontal
- Angle of depression: Looking DOWN from horizontal
These angles are always measured from the horizontal.
Worked Example#
From the top of a 20m building, the angle of depression to a car is $35°$. How far is the car from the base?
The angle of depression from the top equals the angle of elevation from the car (alternate angles, parallel horizontals).
$\tan 35° = \frac{20}{d}$
$d = \frac{20}{\tan 35°} = \frac{20}{0.7002} = 28.6$ m
🚨 Common Mistakes#
- Labelling from the wrong angle: Opposite and Adjacent swap when you change which angle you’re looking from. ALWAYS re-label when the angle changes.
- Calculator in wrong mode: Must be in DEG (degrees), not RAD. Check before every test.
- Inverse trig confusion: $\sin^{-1}$ is NOT $\frac{1}{\sin}$. It means “what angle has this sine value?”
- Forgetting the right angle: SOH CAH TOA only works in RIGHT-ANGLED triangles. For non-right-angled triangles, you need the Sine and Cosine Rules (Grade 11).
- Using Pythagoras when you need an angle: Pythagoras finds sides, not angles. Use inverse trig for angles.
💡 Pro Tip: The “Two-Step” Strategy#
Many problems need two steps:
- Use one ratio (or Pythagoras) to find a missing measurement.
- Use that result in a second ratio to find the final answer.
Always ask: “What do I know? What do I want? What connects them?”
🔗 Related Grade 10 topics:
- Analytical Geometry — gradient = $\tan\theta$, and Pythagoras gives the distance formula
- Solving Equations — trig equations are solved using inverse functions and algebra
📌 Where this leads in Grade 11:
- Reduction Formulas & CAST — extending trig to angles beyond $90°$
- Trig Identities & Equations — proving identities and solving trig equations
- Sine Rule, Cosine Rule & Area Rule — solving ANY triangle, not just right-angled ones