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# 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°

2080 Points
3 years ago
the above question is extremely simple to do, and A=15, B=30 degree
to 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+1-4cos2B
or, 1/2=cos2B=cos60
2B=60 or B=30
from (1), 3sin2A=sin2B=sin60=3/2
or sin2A=1/2=sin30
or, 2A=30
or, A=15 degree