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प्रश्न
If sin (A + B) = 1 and `cos (A - B) = sqrt(3)/2` and 0° < A + B < 90°, A > B then B is equal to ______.
विकल्प
30°
60°
20°
15°
MCQ
रिक्त स्थान भरें
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उत्तर
If sin (A + B) = 1 and `cos (A - B) = sqrt(3)/2` and 0° < A + B < 90°, A > B then B is equal to 30°.
Explanation:
1. sin (A + B) = 1
⇒ A + B = 90° ...(Since sin θ = 1 at θ = 90°)
2. `cos (A - B) = sqrt(3)/2`
⇒ A – B = 30° ...`(cos 30^circ = sqrt(3)/2)`
3. Solve the two equations:
Adding gives 2A = 120°
⇒ A = 60°
Subtracting or solving gives B = 30°.
Hence, B = 30°.
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Notes
The condition 0° < A + B < 90° is likely intended to allow A + B = 90° (i.e., ≤ 90°); otherwise sin (A + B) = 1 would be inconsistent with a strict inequality.
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