In a triangle ABC, \angle A=60^\circ. What is AB^{2}+AC^{2}-BC^{2} equal to?
- A. AB \cdot AC ✓
- B. AB \cdot BC
- C. AC \cdot BC
- D. 2 AB \cdot AC
Correct Answer: A. AB \cdot AC
Explanation
Applying the Cosine Rule: BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A. Since \angle A = 60^\circ and \cos 60^\circ = \frac{1}{2}, this simplifies to BC^2 = AB^2 + AC^2 - AB \cdot AC. Rearranging gives AB^2 + AC^2 - BC^2 = AB \cdot AC.
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