Question

If sin 3φ=cos(φ-6°),where 3φ and (φ-6°) are both acute angles , then find the value of φ

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

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