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\(\tan \angle ACB=\cfrac{AB}{BC}=\cfrac{\sqrt{3}}{1}\)
বা, \(\tan \angle ACB=\tan 60^o\)
বা, \(\angle ACB=60^o\)

আবার, \(\tan \angle BCD=\cfrac{BD}{BC}=\cfrac{1}{1}=1=\tan 45^o\)
বা, \( \angle BCD=45^o\)

\(\therefore \angle ACD=\angle ACB-\angle BCD= 60^o - 45^o=15^o \)