Q1. [2] § 8.3 Trigonometric Ratios of Some Specific Angles § 8.5 Summary
If $\sin(A - B) = \dfrac{1}{2}$, $\cos(A + B) = \dfrac{1}{2}$; $0 < A + B \leq 90°$, $A > B$; find $\angle A$ and $\angle B$.
Previously asked in CBSE board exam
2024 30/4/1 Q23(b) (OR-2)
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Since $\sin(A - B) = \dfrac{1}{2}$, we get $A - B = 30°$ … (1)
Since $\cos(A + B) = \dfrac{1}{2}$, we get $A + B = 60°$ … (2)
Adding (1) and (2): $2A = 90°$ ⟹ A = 45°
Subtracting (1) from (2): $2B = 30°$ ⟹ B = 15°
Source: Chapter 8, Section 8.3, Example 8
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Explanation
- Examiner expects you to use standard trig values: $\sin 30° = \frac{1}{2}$ and $\cos 60° = \frac{1}{2}$ to set up the two equations directly.
- Then solve the simultaneous equations by addition and subtraction — this is the key step that earns both marks.
- Writing both equations clearly and showing the solving step is necessary; just stating A = 45°, B = 15° without working will lose a mark.
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