→ t302850725 :謝謝! 03/05 13:18
※ 引述《t302850725 (MMYX)》之銘言:
: 三角形ABC,分別對應邊長abc,
: 已知abc=35 a+b+c=10
: 角B為60度
: 求外接圓邊長R及b= ??
: 感謝各位
: -----
: Sent from JPTT on my HTC One 801e.
a+b+c=10, abc=35, a^2+c^2-b^2=ac
a, b and c are the roots of x^3-10x^2+(ab+bc+ca)x-35=0.
2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)
=100-(2b^2+ac)
=100-2b^2-35/b
→x^3-10x^2+(50-b^2-35/2b)x-35=0
b is one of the root of the equation
so b^3-10b^2+50b-b^3-35/2-35=0
4b^2-20b+21=0 b=7/2 or 3/2
When b=3/2, a and c are not real numbers.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.112.212.153