How do you calculate #tan(arccos(5/13))#?

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Answer 1 · 2015-02-06T14:50:39

The answer is: #12/5#.

If you set #alpha=arccos(5/13)rArrcosalpha=5/13# than we have to calculate:

#tanalpha=sinalpha/cosalpha#.

#cosalpha=5/13rArrsinalpha=sqrt(1-cos^2alpha)=sqrt(1-25/169)=#

#=sqrt((169-25)/169)=sqrt(144/169)=12/13#.

So:

#tanalpha=(12/13)/(5/13)=12/13*13/5=12/5#.