Vì ABC cân tại A 

\(\Rightarrow\widehat{B}=\widehat{C}\)

Xét tam giác ABC có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(định.lí\right)\)

\(\Rightarrow2\widehat{B}+\widehat{B}+\widehat{B}=180^o\\ \Rightarrow4\widehat{B}=180^o\\ \Rightarrow\widehat{B}=180^o:4\\ \Rightarrow\widehat{B}=45^o\\ \Rightarrow\widehat{C}=\widehat{B}=45^o\\ \Rightarrow\widehat{A}=2\widehat{B}=2.45^o=90^o\)

Đặt \(\widehat{A}=2\widehat{B}=5\widehat{C}=k\)

\(\Rightarrow\hept{\begin{cases}\widehat{A}=k\\\widehat{B}=\frac{k}{2}\\\widehat{C}=\frac{k}{5}\end{cases}}\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=k+\frac{k}{2}+\frac{k}{5}=180^0\)

\(\Rightarrow\frac{17}{10}k=180^0\Leftrightarrow k=\frac{1800}{17}^0\)

\(\Rightarrow\hept{\begin{cases}\widehat{A}=\frac{1800}{17}^0\\\widehat{B}=\frac{900}{17}^0\\\widehat{C}=\frac{360}{17}^0\end{cases}}\)

\(\widehat{A}=180^o:\left(1+2+3\right).1=30^o\)

\(\widehat{A}=180^o:\left(1+2+3\right).2=60^o\)

\(\widehat{A}=180^o:\left(1+2+3\right).3=90^o\)

a: 

Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:

\(\dfrac{a}{\dfrac{1}{6}}=\dfrac{b}{\dfrac{1}{14}}=\dfrac{c}{\dfrac{1}{21}}=\dfrac{a+b+c}{\dfrac{1}{6}+\dfrac{1}{14}+\dfrac{1}{21}}=\dfrac{180}{\dfrac{2}{7}}=630\)

Do đó: a=105; b=45; c=30

Ta có :

 \(\hept{\begin{cases}A=2B\\2C=3B\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{A}{2}=\frac{B}{1}\\\frac{C}{3}=\frac{B}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{A}{4}=\frac{B}{2}\\\frac{C}{3}=\frac{B}{2}\end{cases}\Leftrightarrow}\frac{A}{4}=\frac{B}{2}=\frac{C}{3}}\)

Áp dụng TC của dãy tỉ số bằng nhau , ta có :

\(\frac{A}{4}=\frac{B}{2}=\frac{C}{3}=\frac{A+B+C}{4+2+3}=\frac{180}{9}=20\)

\(\Rightarrow\hept{\begin{cases}A=20.4=80^o\\B=20.2=40^o\\C=20.3=60^o\end{cases}}\)

Ta có: \(\widehat{C}-\widehat{B}=36^0\Rightarrow\widehat{C}=\widehat{B}+36^0\)

Xét tam giác ABC có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

\(\Rightarrow2\widehat{B}+\widehat{B}+\widehat{B}+36^0=180^0\Rightarrow\widehat{B}=36^0\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=2\widehat{B}=72^0\\\widehat{C}=\widehat{B}+36^0=72^0\end{matrix}\right.\)