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A and B are positive acute angles satisfying the equations 3cos2A + 2cos2B = 4 and (3sinA/sinB) = (2cosB/cosA), then A + 2B is equal to A) 45 B) 60 C) 30 D) 90
A and B are positive acute angles satisfying the equations 3cos2A + 2cos2B = 4 and (3sinA/sinB) = (2cosB/cosA), then A + 2B is equal to
A) 45
B) 60
C) 30
D) 90
3cos2A + 2cos2B = 4 ................1
cos2A = 2cos2A - 1
cos2B = 2cos2B - 1
eq 1 becomes
3(1+cos2A)/2 + (1+cos2B) = 4 .................2
3sinA/sinB = 2cosB/cosA
3sinAcosA =2sinB cosB (sin2@ = 2sin@cos@)
3sin2A = 2sin2B .......................3
squaring eq 3
9sin22A = 4sin22B
9(1-cos22A) = 4(1-cos22B)
9cos22A - 4cos22B = 5 ....................4
after solving 4 & 2 we get
cos2A = 7/9 & cos2B = 1/3
cos2A = 2cos2A-1 so
cosA = 2root2/3 , cos2B = 1/3
A + 2B = cos-12root2/3 + cos-11/3 (cos-1x+cos-1y = cos-1{xy-[(1-x2)(1-y2)]1/2}
=cos-10 = pi/2 = 90o ...
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