\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \Rightarrow2\widehat{B}+15^0+\widehat{C}=180^0\\ \Rightarrow2\widehat{C}+60^0+15^0+\widehat{C}=180^0\\ \Rightarrow3\widehat{C}=105^0\Rightarrow\widehat{C}=35^0\\ \Rightarrow\widehat{B}=65^0\\ \Rightarrow\widehat{A}=80^0\)
Ta có: \(\left\{{}\begin{matrix}\widehat{B}+15^0=\widehat{A}\\\widehat{C}+30^0=\widehat{B}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=\widehat{B}+15^0\\\widehat{C}=\widehat{B}-30^0\end{matrix}\right.\)
Xét tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(tổng 3 góc trong tam giác )
\(\Rightarrow\widehat{B}+15^0+\widehat{B}+\widehat{B}-30^0=180^0\)
\(\Rightarrow3\widehat{B}=195^0\Rightarrow\widehat{B}=65^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=\widehat{B}+15^0=65^0+15^0=80^0\\\widehat{C}=\widehat{B}-30^0=65^0-30^0=35^0\end{matrix}\right.\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
Thay các số đo ở đề, ta được: \(\left(\widehat{C}+30^o\right)+\left(\widehat{C}+30^o+15^o\right)+\widehat{C}=180^o\)
Giải ra, ta được: \(\widehat{C}=35^o\)
=> \(\widehat{B}=35^o+30^o=65^o\)
\(\widehat{A}=65^o+15^o=80^o\)
