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Given that A,B are positive acute angles and √3sin2A=sin2B and √3sin 2 A+sin 2 B=(√3-1)÷2 Then A or B may take the values- A. 15° B. 30° C. 45° D. 75°
Given that A,B are positive acute angles and √3sin2A=sin2B and √3sin2A+sin2B=(√3-1)÷2Then A or B may take the values-A. 15°B. 30°C. 45°D. 75°

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2 years ago

```							the above question is extremely simple to do, and A=15, B=30 degreeto solve this, simply write sin^2A and sin^2B in the second equation in terms of cos2A and cos2B (using the formula cos2x=1-2sin^2x).so, the second equation becomes: √3cos2A=2-cos2B   …...(2)while the first equation is given as:√3sin2A=sin2B …...(1)squaring and adding both the equations, we get: 3= 4+1-4cos2Bor, 1/2=cos2B=cos602B=60 or B=30from (1), √3sin2A=sin2B=sin60=√3/2or sin2A=1/2=sin30or, 2A=30or, A=15 degree
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2 years ago
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