Given that sin (5x - 28)^o = cos (3x - 50)^o,0 < x < 90^o, find the value of x

14^o

21^o

32^o

39^o

Correct answer is B

Sin (5x – 28)^o = cos (3x - 50)^o
Since by the trigonometry relation
Sin(5x – 28)^o = cos[90 – (5x – 28)]^o
Hence cos(3x – 50)^o = cos[90 – (5x – 28)]^o
3x – 50 = 90 - (5x-28)
3x – 50 = 90 – 5x + 28
3x + 5x = 90 + 28 + 50
8x = 168
\(x = \frac{168}{8}=21^{\circ}\)

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Given that sin (5x - 28)o = cos (3x - 50)o,0 < x < 90o, find the value of x

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Given that sin (5x - 28)^o = cos (3x - 50)^o,0 < x < 90^o, find the value of x