Using Cos's theorem:

We have : \(BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot cos\angle A\)

\(\Leftrightarrow5,042^2=4,123^2+7,145^2-2\cdot4,123\cdot7,415\cdot cos\angle A\)

\(\Rightarrow\angle A\simeq45,8^o\)

So on : \(AB^2=BC^2+AC^2-2\cdot BC\cdot AC\cdot cos\angle C\)

\(\Rightarrow\angle C\simeq32^o\)

Next : \(AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot cos\angle B\)

\(\Rightarrow\angle B\simeq102,2^o\)

Go to this source: http://tailieu.tv/tai-lieu/chuyen-de-toan-casio-27004/

P/s: Vietnamese is permitted in Mathu.