a) Góc ngoài tại đỉnh A gọi là \(\widehat{A_1}\) có số đo bằng:
\(\widehat{A_1}=\widehat{B}+\widehat{C}=3\widehat{B}\Rightarrow\widehat{C}=2\widehat{B}\) nên \(\widehat{B}=\widehat{\dfrac{C}{2}}\)
\(\widehat{C}=\dfrac{4}{3}\widehat{A}\Rightarrow\widehat{A}=\dfrac{3}{4}\widehat{C}\)
Tổng các góc trong tam giác bằng 1800, ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Rightarrow\dfrac{3}{4}\widehat{C}+\widehat{\dfrac{C}{2}}+\widehat{C}=180^0\)
\(\dfrac{9}{4}\widehat{C}=180^0\Rightarrow\widehat{C}=\dfrac{4.180^0}{9}=80^0\)
\(\widehat{B}=\dfrac{80^0}{2}=40^0;\widehat{A}=\dfrac{3}{4}.80^0=60^0\)
b) Gọi góc ngoài tại đỉnh C là \(\widehat{C_1}=\widehat{A}+\widehat{B}=4\widehat{B}\)
\(3\widehat{B}=\widehat{A}\Rightarrow\widehat{B}=\widehat{\dfrac{A}{3}}\)
\(\widehat{A}-\widehat{C}=100^0\Rightarrow\widehat{C}=\widehat{A}-100^0\)
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Rightarrow\widehat{A}+\widehat{\dfrac{A}{3}}+\widehat{A}-100^0=180^0\)
\(\Rightarrow\dfrac{7\widehat{A}}{3}=280^0\Rightarrow\widehat{A}=\dfrac{3.280^0}{7}=120^0\)
\(\Rightarrow\widehat{B}=\dfrac{\widehat{A}}{3}=40^0;\widehat{C}=\widehat{A}-100^0=20^0\)
c)
\(\left\{{}\begin{matrix}\widehat{A}-\widehat{B}=45^0\\\widehat{A}-\widehat{C}=30^0\\\widehat{A}+\widehat{B}+\widehat{C}=180^0\end{matrix}\right.\Rightarrow3\widehat{A}=225^0\)
\(\Rightarrow\) \(\widehat{A}=225^0:3=85^0\\ \widehat{B}=85^0-45^0=40^0\\ \widehat{C}=85^0-30^0=55^0\)
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