\(\Delta ABC\) có
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
Mà theo GT ta có:
\(\widehat{A}=\frac{2}{3}\widehat{C}\) và \(\widehat{B}=55^0\)
=> \(\frac{2}{3}\widehat{C}+55^0+\widehat{C}=180^0\)
\(\left(\frac{2}{3}+1\right).\widehat{C}+55^0=180^0\)
\(\frac{5}{3}\widehat{C}+55^0=180^0\)
\(\frac{5}{3}\widehat{C}=180^0-55^0=125^0\)
\(\widehat{C}=125^0:\frac{5}{3}=125^0.\frac{3}{5}=75^0\)
Vì \(\widehat{A}=\frac{2}{3}\widehat{C}\Rightarrow\widehat{A}=\frac{2}{3}.75^0=50^0\)
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