Quandaries and Queries


Name:  Kyle
Student, Secondary Question
I would like to know how to evaluate the problem of:  Arccos 5/13.
Thank you.


Hi Kyle,

I expect that what you want is some trig function of arccos(5/13), like maybe


I would start by drawing a triangle.

In this diagram the angle ABC has a cosine of 5/13. In other words

arccos(5/13) = angle ABC

By Pythagoras' theorem

52 + |CB|2 = 132

that is

|CA|2 = 132 - 52 = 169 - 25 = 144 so
|CA| = 12

Now you can read the trig functions of angle ABC from the diagram, for example

tan(arccos(5/13)) = 12/5

I hope this helps,


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