Question

if sin3A=cos(A-6) where (3A) and (A-6) are both acute angle,then find the value of A

Answer 1

sin 3A = cos (A-6)  sin 3A = sin (90-(A-6))   ( since sin(90- A) = cos A)  therefore, 3A = 90-(A-6)  3A = 90-A+6  3A+A = 96  4A = 96  A = 96/4  A = 24  Please mark as brainliest answer

Answer 2

sin3A=cos[A-6] sin3A=sin(90-(A-6) therefor 3a=90-A+6 3a+a= 96 so 4a =96 a=96/4= 24