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
Since AK is an altitude, triangles AKC and AKB are right
AK = sqrt [ AC^2 - CK^2 ]
AK = sqrt [ 8^2 - 3^2] = sqrt [ 64 - 9 ] = sqrt 55
AB = sqrt [ AK^2 + BK^2 ]
AB = sqrt [ 55 + 2^2 ] = sqrt [59]
+380 
