a)

 

=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180^o

100^o + \(\widehat{B}+\widehat{C}\) = 180^o

\(\widehat{B}+\widehat{C}\) = 180^o - 100^o

\(\widehat{B}+\widehat{C}\) = 80^o

Góc B = (80^o+50^o):2 = 65^o

=> \(\widehat{C}\) = 65^o - 50^o = 15^o

Vậy \(\widehat{B}\) = 65^o ; \(\widehat{C}\) = 15^o

b)

 

Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180^o

\(\widehat{3A}+\widehat{2C}\) = 180^o - 80^o

\(\widehat{3A}+\widehat{2C}\) = 100^o

=> \(\widehat{A}\) = 100^o:(3+2).3 = 60^o

\(\widehat{C}\) = 100^o - 60^o = 40^o

Vậy \(\widehat{A}\) = 60^o ; \(\widehat{C}\) = 40^o

Chọn A

A

A

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

Cảm ơn bạn nhiều!!

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 \(\widehat{C}>0\)

$\widehat{ABC}$

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 \(\widehat{B}=75^0;\widehat{C}=25^0\)