Asked by T
what is the cot of theta if the cos of theta = 4/9 and theta lies in quadrant IV
Answers
Answered by
Reiny
cosØ = 4/9 = adj/hyptenuse of the triangle in quadrant IV
4^2 + y^2 = 81
y = -√65 in quadrant IV
tan Ø = y/x = -√65/4
cot Ø = -4/√65
4^2 + y^2 = 81
y = -√65 in quadrant IV
tan Ø = y/x = -√65/4
cot Ø = -4/√65
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