Question

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

• A. \( 45^\circ \)
• B. \( 60^\circ \)
• C. \( 90^\circ \)
• D. \( 120^\circ \)

Hint:

If \( \cos \theta = 0 \), then \( \theta = 90^\circ \). This means the given triangle is a right-angled triangle.

Answer 1

The Correct Option is C

To determine the angle \( B \), we use the **Cosine Rule**: \[ \cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} \] Substituting the given values: \[ \cos B = \frac{(6\sqrt{3})^2 + 6^2 - 12^2}{2 \times (6\sqrt{3}) \times 6} \] \[ = \frac{108 + 36 - 144}{2 \times 6\sqrt{3} \times 6} \] \[ = \frac{0}{72\sqrt{3}} \] \[ = 0 \] Since \( \cos B = 0 \), we get: \[ B = 90^\circ \]