fie AD⊥BC, D∈BC
mas∡C=180°-75°-60°=45°
si mas ∡(BAD)=90°-60°=30°
⇄DC=AD=(24√6/√2=24√3
⇄BD=24√3/√3=24 ..am folosit ctg 60°=1/√3
⇆AB=24*2=48 ...am folosit teo unghiului de 30 ° sau am aplicat cos 60° in tr ABD
deci avem
AB=48
AC=24√6
BC-24+24√3
Perimetrul= AB+AC+BC= 48+24√6+24√3+24=
=24(2+√6+√3)