Cho tam giác ABC có góc A = 3 lần góc B; góc A/3 = góc C/5
Tính các góc tam giác ABC
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
\(\widehat{B}=60^0;\widehat{C}=20^0\)
Xét ΔABC có \(\widehat{A}>\widehat{B}>\widehat{C}\)
nên BC>AC>AB
Ta có: \(\hat{A}=2\cdot\hat{C}\)
=>\(\hat{C}=\frac12\cdot\hat{A}\)
TA có: \(\hat{A}=3\cdot\hat{B}\)
=>\(\hat{B}=\frac13\cdot\hat{A}\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}+\frac12\cdot\hat{A}+\frac13\cdot\hat{A}=180^0\)
=>\(\frac{11}{6}\cdot\hat{A}=180^0\)
=>\(\hat{A}=180^0:\frac{11}{6}=\frac{1080^0}{11}\)
=>\(\hat{C}=\frac12\cdot\frac{1080^0}{11}=\frac{540^0}{11};\hat{B}=\frac13\cdot\frac{1080^0}{11}=\frac{360^0}{11}\)
Bài 1:
1. Ta có ^B+^C=1800-1000=800. => ^C=[(^B+^C)-(^B-^C)]/2 =(800-500)/2=150 => ^B=150+500=650.
2. ^A+^C=1800-^B=1800-800=1000
3^A=2^C => ^A/2=^C/3 = (^A+^C)/2+3 (Dãy tỉ số bằng nhau)
=(^A+^C)/5=1000/5=200 => ^A=200.2=400; ^C=200.3=600.
Bài 2:
Gọi góc ngoài đỉnh C của tam giác ABC là ^ACy => ^Cx là phân giác ^ACy
=> ^ACx=^xCy=^ACy/2=1200/2=600
^A=600 => ^ACy=^A=600. Mà 2 góc này so le trong => Cx//AB.