Question

In \( \triangle ABC \), \( AB = 6\sqrt{3} \, \text{cm}, AC = 12 \, \text{cm}, \) and \( BC = 6 \, \text{cm} \), then \( \angle B \) is

• A. 45°
• B. 60°
• C. 90°
• D. 120°

Hint:

Use the cosine rule to find angles in triangles when the side lengths are known.

Answer 1

The Correct Option is C

Step 1: Using the given values \( AB = 6\sqrt{3} \, \text{cm}, AC = 12 \, \text{cm}, BC = 6 \, \text{cm} \), we can apply the cosine rule to find \( \angle B \). The cosine rule states: \[ \cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} \] Step 2: Substituting the values: \[ \cos B = \frac{(6\sqrt{3})^2 + 6^2 - 12^2}{2 \times (6\sqrt{3}) \times 6} \] \[ \cos B = \frac{108 + 36 - 144}{72\sqrt{3}} = 0 \] Thus, \( \angle B = 90^\circ \).