If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y.... Question 23430

If sin 3y = cos 2y and 0^o \(\leq\) 90^o, find the value of y

18^o

36^o

54^o

90^o

Correct answer is A

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = \(\frac{90}{5}\)

y = 18^o

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