We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude to $\triangle ABC$. If $AC = 8,$ $BK = 2$, and $CK = 3,$ then what is $AB$?
A
8
B 2 K 3 C
AK = sqrt (AC^2 - CK^2 ) = sqrt (8^2 - 3^2 ) = sqrt (55)
AB = sqrt (BK^2 + AK^2 )
AB = sqrt ( 2^2 + 55)
AB = sqrt [ 59 ] ≈ 7.68
+585 
