a) \(\widehat{A}=70^0\Rightarrow\widehat{B}+\widehat{C}=180-70=110^0\)
\(\widehat{B}-\widehat{C}=10^0\)
=>\(\widehat{B}=\frac{110+10}{2}=60^0\)
\(\Rightarrow\widehat{C}=110-60=50^0\)
b) \(\widehat{A}=100^0\Rightarrow\widehat{B}+\widehat{C}=180-100=80^0\)
\(\widehat{B}-\widehat{C}=50^0\)
\(\Rightarrow\widehat{B}=\frac{80+50}{2}=65^0\)
\(\Rightarrow C=80-65=15^0\)
c) \(\widehat{A}=60^0\Rightarrow\widehat{B}+\widehat{C}=180=60=120^0\)
\(\widehat{B}=2\widehat{C}\)
\(\Rightarrow\widehat{B}=\frac{120}{3}.2=80^0\)
\(\Rightarrow C=120-80=40^0\)
a. Ta có: \(\widehat{B}+\widehat{C}=180^o-70^o=110^o\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\frac{110^o+10^o}{2}=60^o\\\widehat{C}=110^o-60^o=50^o\end{matrix}\right.\)
b. Ta có: \(\widehat{B}+\widehat{C}=180^o-100^o=80^o\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\frac{80^o+50^o}{2}=65^o\\\widehat{C}=80^o-65^o=15^o\end{matrix}\right.\)
c. Ta có: \(\widehat{A}+2\widehat{C}+\widehat{C}=180^o\Rightarrow60^o+3\widehat{C}=180^o\Rightarrow\widehat{C}=40^o\)
\(\Rightarrow\widehat{B}=2\widehat{C}=2.40^o=80^o\)
a, \(\widehat{B}-\widehat{C}=10^o\Rightarrow\widehat{B}=10^o+\widehat{C}\)
Xét △ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow70^o+10^o+2\widehat{C}=180^o\\ \Rightarrow2\widehat{C}=100^o\\ \Rightarrow\widehat{C}=50^o\\ \Rightarrow\widehat{B}=50^o+10^o=60^o\)
b, \(\widehat{B}-\widehat{C}=50^o\Rightarrow\widehat{B}=50^o+\widehat{C}\)
Xét △ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow100^o+50^o+2\widehat{C}=180^o\\ \Rightarrow2\widehat{C}=30^o\\ \Rightarrow\widehat{C}=15^o\\ \Rightarrow\widehat{B}=50^o+15^o=65^o\)
c) Xét △ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow60^o+3\widehat{C}=180^o\\ \Rightarrow3\widehat{C}=120^o\\ \Rightarrow\widehat{C}=40^o\\ \Rightarrow\widehat{B}=2.40^o=80^o\)
a/ ΔABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=> \(\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-70^0=110^0\)
Ta có: \(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=110^0\\\widehat{B}-\widehat{C}=10^0\end{matrix}\right.\)
Cộng vế theo vế ta có:
\(2.\widehat{B}=120^0\)
=> \(\widehat{B}=120^0:2=60^0\)
\(\widehat{B}+\widehat{C}=110^0\)
=> \(\widehat{C}=110^0-\widehat{B}=110^0-60^0=50^0\)
Vậy:....................
b/ ΔABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=> \(\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-100^0=80^0\)
Ta có: \(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=80^0\\\widehat{B}-\widehat{C}=50^0\end{matrix}\right.\)
Cộng vế theo vế ta có:
\(2.\widehat{B}=130^0\)
=> \(\widehat{B}=130^0:2=75^0\)
\(\widehat{B}-\widehat{C}=50^0\)
=> \(\widehat{C}=\widehat{B}-50^0=75^0-50^0=25^0\)
