In the right-angled triangle ABC, `angleB=90^(@)`.
`therefore angleACB+angleBAC=90^(@) …………………(1)`
`tan angleACB=(AB)/(BC)[tan theta=("perpendicular")/("base")]`
`=(8sqrt(3))/(8)`
`=sqrt(3)=tan60^(@)`
`therefore angleACB=60^(@)`
from (1) we get, `angleBAC=90^(@)-angleACB=90^(@)-60^(@)=30^(@)`.
Hence the required value of `angleACB=60^(@) and angleBAC=30^(@).`