Given that Sin (5\(_x\) − 28)\(^o\) = Cos(3\(_x\) − 50)\(^o\), o < x < 90\(^o\)
Find the value of x

A.

14\(^o\)

B.

21\(^o\)

C.

32\(^o\)

D.

39\(^o\)

Correct answer is B

Sin(5x - 28) = Cos(3x - 50)………..i

  But Sinα = Cos(90 - α)

  So Sin(5x - 28) = Cos(90 - [5x - 28])

  Sin(5x - 28) = Cos(90 - 5x + 28)

  Sin(5x - 28) = Cos(118 - 5x)………ii

  Combining i and ii

  Cos(3x - 50) = Cos(118 - 5x)

  3x - 50 = 118 - 5x

  Collecting the like terms

  3x + 5x = 118 + 50

  8x = 168

  x = \(\frac{168}{8}\)

  x = 21\(^o\)

  Answer is B