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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 toA) 45B) 60C) 30D) 90

10 years ago

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 ...