Mathematics — CBSE Class 10 board question

Given: $2\sin(A+B) = \sqrt{3}$ and $\cos(A-B) = 1$

From the first equation:
$$\sin(A+B) = \frac{\sqrt{3}}{2} \Rightarrow A+B = 60° \quad \cdots(1)$$

From the second equation:
$$\cos(A-B) = 1 = \cos 0° \Rightarrow A-B = 0° \quad \cdots(2)$$

Adding (1) and (2): $2A = 60° \Rightarrow \mathbf{A = 30°}$

Subtracting (2) from (1): $2B = 60° \Rightarrow \mathbf{B = 30°}$

Source: Chapter 8, Section 8.3

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• Examiners expect you to first convert each equation to a standard angle using the trig table (sin 60° = √3/2, cos 0° = 1), then solve the simultaneous equations. Writing both equations clearly and showing the addition/subtraction step earns full marks.
• Note: cos(A−B) = 1 directly gives A−B = 0°, since cos 0° = 1 is a standard value from the table.