Mathematics — CBSE Class 10 board question

From $\sin(A - B) = 0$:
$$A - B = 0° \quad \Rightarrow \quad A = B \quad \cdots (1)$$

From $2\cos(A + B) - 1 = 0$:
$$\cos(A + B) = \frac{1}{2} \quad \Rightarrow \quad A + B = 60° \quad \cdots (2)$$

Solving (1) and (2):

$$2A = 60° \quad \Rightarrow \quad A = 30°, \quad B = 30°$$

∴ A = 30° and B = 30°

Source: Chapter 8, Section 8.3 (Example 8 pattern)

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• $\sin(A-B) = 0$ means $A - B = 0°$ (since $\sin 0° = 0$).
• $2\cos(A+B) = 1 \Rightarrow \cos(A+B) = \frac{1}{2} \Rightarrow A+B = 60°$ (since $\cos 60° = \frac{1}{2}$).
• Add the two equations to get A, then subtract to get B. Show both steps clearly — examiners award one mark for setting up the equations correctly and one mark for the final values.