In a right-angled triangle ABC, sin(A+B) = 1 and cos A = \(\frac{1}{2}\), find value of cos (C-B)?
Answer & explanation
Correct answer: option 1
sin(A+B) = 1 ⇒ A+B = 90°
cosA = \(\frac{1}{2}\) ⇒ A = 60°
⇒ B = 30°
⇒ A + B + C = 180°
⇒ (60° + 30°) + C = 180°
⇒ C = 90°
∴ cos (C-B) = cos (90° - 30°) = cos60° = \(\frac{1}{2}\)