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?

Since 16² + 63² = 256 + 3969 = 4225 = 65², the triangle is right-angled at B (the angle opposite to the hypotenuse AC = 65). Hence B = 90°, so 2B = 180° and cos 2B = –1. Also, A + C = 90°, so 2A + 2C = 180°, giving cos 2A = –cos 2C. Therefore cos 2A + cos 2C = 0. Adding, cos 2A + cos 2B + cos 2C = 0 + (–1) = –1.