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Ta có: \(\widehat{C}-\widehat{B}=36^0\Rightarrow\widehat{C}=\widehat{B}+36^0\)
Xét tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
\(\Rightarrow2\widehat{B}+\widehat{B}+\widehat{B}+36^0=180^0\Rightarrow\widehat{B}=36^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=2\widehat{B}=72^0\\\widehat{C}=\widehat{B}+36^0=72^0\end{matrix}\right.\)
Lời giải:
a. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{A}{3}=\frac{B}{4}=\frac{C}{5}=\frac{A+B+C}{3+4+5}=\frac{180^0}{12}=15^0$
$\Rightarrow A=3.15^0=45^0; B=4.15^0=60^0; C=5.15^0=75^0$
b. Áp dụng TCDTSBN:
$A=2B=6C$
$= A=\frac{B}{\frac{1}{2}}=\frac{C}{\frac{1}{6}}$
$=\frac{A+B+C}{1+\frac{1}{2}+\frac{1}{6}}=\frac{180^0}{\frac{5}{3}}=108^0$
$\Rightarrow A=108^0; B=108^0.\frac{1}{2}=54^0; C=108^0.\frac{1}{6}=18^0$
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
a) \(\widehat{A}=2\widehat{B}=6\widehat{C}\Leftrightarrow\frac{\widehat{A}}{2}=\frac{\widehat{B}}{6}=\frac{\widehat{C}}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{\widehat{A}}{2}=\frac{\widehat{B}}{6}=\frac{\widehat{C}}{1}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{2+6+1}=\frac{180^o}{9}=20^o\)
\(\frac{\widehat{A}}{2}=20^o\Rightarrow\widehat{A}=20^o.2=40^o\)
\(\frac{\widehat{B}}{6}=20^o\Rightarrow\widehat{A}=20^o.6=120^o\)
\(\widehat{C}=20^o\)
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b,d tương tự
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