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a)
=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180o
100o + \(\widehat{B}+\widehat{C}\) = 180o
\(\widehat{B}+\widehat{C}\) = 180o - 100o
\(\widehat{B}+\widehat{C}\) = 80o
Góc B = (80o+50o):2 = 65o
=> \(\widehat{C}\) = 65o - 50o = 15o
Vậy \(\widehat{B}\) = 65o ; \(\widehat{C}\) = 15o
b)
Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180o
\(\widehat{3A}+\widehat{2C}\) = 180o - 80o
\(\widehat{3A}+\widehat{2C}\) = 100o
=> \(\widehat{A}\) = 100o:(3+2).3 = 60o
\(\widehat{C}\) = 100o - 60o = 40o
Vậy \(\widehat{A}\) = 60o ; \(\widehat{C}\) = 40o
Ra rồi đây.
Ta có: \(\widehat{B}+\widehat{C}+\widehat{A}=180\) độ
\(\Rightarrow4\widehat{A}+4\widehat{A}+\widehat{A}=180\)độ
\(\Rightarrow9\widehat{A}=180\Rightarrow\widehat{A}=180:9=20\)độ
Vì \(\widehat{A}-\widehat{B}=\widehat{B}-\widehat{C}\) nên \(\widehat{A}-2\widehat{B}+\widehat{C}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}-2\widehat{B}+\widehat{C}=0^0\left(1\right)\\\widehat{A}+\widehat{B}+\widehat{C}=180^0\left(2\right)\end{matrix}\right.\)
Trừ \(\left(2\right)\) cho \(\left(1\right)\), ta được \(3\widehat{B}=180^0\Rightarrow\widehat{B}=60^0\)
\(\Rightarrow\widehat{A}+\widehat{C}=120^0\)
Vậy GTLN của \(\widehat{A}\) là \(119^0\) vì \(\widehat{C}>0\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (tổng ba góc của một tam giác)
\(\Rightarrow\widehat{B}+\widehat{C}=180^0-80^0\)
\(\Rightarrow\widehat{B}+\widehat{C}=100^0\)
Theo đề bài: \(\widehat{B}=3\widehat{C}\)
\(\Rightarrow3\widehat{C}+\widehat{C}=100^0\)
\(\Rightarrow4\widehat{C}=100^0\)
\(\Rightarrow\widehat{C}=25^0\)
\(\Rightarrow\widehat{B}=3\widehat{C}=3.25^0=75^0\)
Vậy \(\widehat{B}=75^0;\widehat{C}=25^0\)
Ta có : \(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-75^o=105^o\)
a/ \(\widehat{B}=2\widehat{C}\Rightarrow2\widehat{C}+\widehat{C}=105^o\Rightarrow3\widehat{C}=105^o\Rightarrow\widehat{C}=35^o\Rightarrow\widehat{B}=70^o\)
b/ \(\widehat{B}-\widehat{C}=25^o\Rightarrow\widehat{B}=\widehat{C}+25^o\Rightarrow\widehat{C}+25^o+\widehat{C}=105^o\Rightarrow2\widehat{C}=80^o\Rightarrow\widehat{C}=40^o\Rightarrow\widehat{B}=65^o\)