Given that Sin (5\(_x\) &minus; 28)\(^o\) = Cos(3\(_x\) &minus; 50)\(^o\), o &lt; x &lt; 90\(^o\)...

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

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

There is an explanation video available below.

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