In a triangle ABC, AB=16 cm, BC=63 cm and AC=65 cm. What is the value of \cos 2A+\cos 2B+\cos 2C?
- A. -1 ✓
- B. 0
- C. 1
- D. \frac{76}{65}
Correct Answer: A. -1
Explanation
Check the sides using the Pythagorean theorem: 16^2 + 63^2 = 256 + 3969 = 4225 = 65^2. This means \triangle ABC is right-angled at B, so B = 90^{\circ} and A+C = 90^{\circ}. Thus, \cos 2B = \cos 180^{\circ} = -1. Since 2C = 180^{\circ} - 2A, we get \cos 2C = -\cos 2A. The expression becomes \cos 2A - 1 - \cos 2A = -1.