1-cos^2x=0.8725
cos^2(x) = .1575
cosx = ± .35707
since x is acute, it must be in the first quadrant, so just take the inverse cosine of .35707 to get
x = 69.1º
I asked this question yesterday, but I am still having alot of trouble, can someone please help me. Thanks.
If 1-cos^2x=0.8725, and x is acute, determine the value of x correct to the nearest tenth of a degree.
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