WAEC - Mathematics (2010 - No. 16)
If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y
18o
36o
54o
90o
Explanation
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 = 18o
sin 3y = cos(90 - 3y)
cos(90 - 3y) = cos 2y
90 - 3y = 2y
5y = 90
y = \(\frac{90}{5}\)
y = 18o
Comments (0)
