In ABC, right-angled at B, AB = 5 cm and ACB = 30° (see Fig. 8.19). Determine the lengths of the sides BC and AC.
Step-by-Step Solution
Key Concept: Use the trigonometric ratios for a right‑angled triangle: \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\theta = 30^{\circ}\) we have \(\sin 30^{\circ}=\frac{1}{2}\) and \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\). The given side AB is opposite \(30^{\circ}\).
1. Identify the known parts of the triangle:\\
- \(\angle B = 90^{\circ}\) (right angle)
- \(\angle C = 30^{\circ}\)
- \(\angle A = 180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}\)
- Side \(AB = 5\,\text{cm}\) is opposite \(\angle C\).\
2. Apply the definition of sine for \(\angle C\):
$$\sin 30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC}$$\
Since \(\sin 30^{\circ}=\frac{1}{2}\),
$$\frac{1}{2}=\frac{5}{AC}\quad\Rightarrow\quad AC = 10\,\text{cm}.$$\
3. Apply the definition of cosine for \(\angle C\):
$$\cos 30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AC}$$\
With \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\) and \(AC=10\) cm,
$$\frac{\sqrt{3}}{2}=\frac{BC}{10}\quad\Rightarrow\quad BC = 10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\,\text{cm}.$$\
4. Hence the required lengths are:
- \(BC = 5\sqrt{3}\,\text{cm}\)
- \(AC = 10\,\text{cm}\).
Correct Answer: BC = 5\sqrt{3}\,\text{cm}, \; AC = 10\,\text{cm}