AA'⊥(ABC)
B∈(ABC) din cele 2 ⇒proiectia lui A'Bpe (ABC)=A'B⇒m∡(A"B,(ABC))=
=m∡(A'B,AB)=m∡(A'BA)
dar tg∡(A'BA)=4√3/4=√3⇒m∡(A'BA)=60°
altfel
A'B=√(AA'²+AB²)=√((4√3)²+4²)=8⇒m∡(A'AB)=30° (teorema unghiuluide 30° in tr dr A'AB)⇒m∡(A'BA)=90°-30°=60°
as simple as that!!