For a right-angled triangle with angle $\theta$:
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$
Mnemonic: SOH–CAH–TOA
| Ratio | Use when you have… |
|---|---|
| $\sin$ | opposite and hypotenuse |
| $\cos$ | adjacent and hypotenuse |
| $\tan$ | opposite and adjacent |
The sides depend on which angle $\theta$ you are using:
- Hypotenuse: opposite the right angle (always the longest side)
- Opposite: the side directly opposite angle $\theta$
- Adjacent: the side next to angle $\theta$ (not the hypotenuse)
$$x = \text{hypotenuse} \times \sin\theta \quad \text{or} \quad x = \text{hypotenuse} \times \cos\theta \quad \text{etc.}$$
In a right-angled triangle, the hypotenuse is 15 m and one angle is $35°$. Find the opposite side.
$$\sin(35°) = \frac{\text{opposite}}{15} \implies \text{opposite} = 15\sin(35°) \approx 15 \times 0.5736 \approx 8.60 \text{ m}$$
Use the inverse trigonometric functions: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$.
$$\theta = \sin^{-1}!\left(\frac{\text{opp}}{\text{hyp}}\right)$$
A ramp rises 1.2 m over a horizontal distance of 5.5 m. Find the angle of inclination.
$$\tan\theta = \frac{1.2}{5.5} \implies \theta = \tan^{-1}!\left(\frac{1.2}{5.5}\right) \approx \tan^{-1}(0.2182) \approx 12.3°$$
Both are measured from the horizontal, not the vertical.
From a cliff 80 m high, the angle of depression to a boat at sea is $22°$. Find the horizontal distance to the boat.
$$\tan(22°) = \frac{80}{d} \implies d = \frac{80}{\tan(22°)} \approx \frac{80}{0.4040} \approx 198 \text{ m}$$
Bearings are measured clockwise from North and written as three digits, e.g., 045°, 270°.
REMEMBER: Always draw a diagram. Label the right angle, the known angle $\theta$, and the known side. Then select the correct ratio (SOH-CAH-TOA).
EXAM TIP: Set your CAS to degree mode for all trigonometry problems in General Mathematics. Check: $\sin(30°)$ should give 0.5.
