Question

Cho tam giác ABC biết 3A+C=155 Độ ,2B-3A=75 Độ .Tính mỗi góc của tam giác ABC

Answer 1

\(3\cdot\widehat{A}+\widehat{C}=155^0\)

=>\(\widehat{C}=155^0-3\cdot\widehat{A}\)

\(2\widehat{B}-3\cdot\widehat{A}=75^0\)

=>\(2\cdot\widehat{B}=3\cdot\widehat{A}+75^0\)

=>\(\widehat{B}=\dfrac{3}{2}\cdot\widehat{A}+37,5^0\)

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{A}+155^0-3\cdot\widehat{A}+\dfrac{3}{2}\cdot\widehat{A}+37,5^0=180^0\)

=>\(-\dfrac{1}{2}\cdot\widehat{A}=180^0-155^0-37,5^0=-\dfrac{25}{2}\)

=>\(\widehat{A}=25^0\)

\(\widehat{B}=\dfrac{3}{2}\cdot25^0+37,5^0=75^0\)

\(\widehat{C}=155^0-3\cdot25^0=80^0\)