# What is cos A given AB=5,AC=6,BC=7?

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You need to use the law of cosines such that:

`(BC)^2 = (AB)^2 + (AC)^2 - 2AB*AC*cos hat A`

`2AB*AC*cos hat A = (AB)^2 + (AC)^2 - (BC)^2`

`cos hat A = ((AB)^2 + (AC)^2 - (BC)^2)/(2AB*AC)`

You need to substitute 5 for AB, 6 for AC, 7 for BC such that:

`cos hat A = (25 +36 - 49)/(60)`

`cos hat A = (61-49)/60 => cos hat A = 12/60 => cos hat A = 1/5`

**Hence, evaluating the cos hat A using the law of cosines yields `cos hat A = 1/5` .**

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