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本帖最后由 王守恩 于 2021-12-1 06:41 编辑
\(好题!简单!ACD是一个等腰直角三角形。记CO=1\ \ ∠CAD=2a\ \ ∠CAB=2b\ \ ∠COB=c\)
\(解方程:\frac{OA\sin(2a)}{\sin(c+2a)}=\frac{\sin(a)}{\sin(c-a)}=\frac{\cos(b)}{\cos(c+b)},
\frac{\sin(a)\sin(2a)\cos(3b)}{\cos(b)]\sin(2b)\sin(3a)}=\frac{\cos(a)\sin(2b)\cos(c+b)\sin(c+2a)}{\cos(b)\sin(2a)\sin(c-a)\sin(c-2b)}=1\)
\(OA=2,\ \ a=\pi/4,\ \ 兼得BE=2,\ AD=3,\ BC=\sqrt{10},\ BO=OD=\sqrt{13},\ CD=\sqrt{18}\)
Solve[{((OA) Sin[2 a])/Sin[c + 2 a] == Sin[a]/Sin[c - a] == Cos[ b]/Cos[c + b],
(Sin[a] Sin[2 a] Cos[3 b])/(Cos[ b] Sin[2 b] Sin[3 a]) ==
( Cos[a] Sin[2 b] Cos[c + b] Sin[c + 2 a])/(Cos[ b] Sin[2 a] Sin[c - a] Sin[c - 2 b]) == 1,
\[Pi]/2 > a > 0, \[Pi]/2 > b > 0, \[Pi] > c > 0, OA > 0}, {OA, a, b, c}] // FullSimplify
{{OA -> 2, a -> \[Pi]/4, b -> -2 ArcTan[3 - Sqrt[10]], c -> 2 ArcTan[1/3 (-2 + Sqrt[13])]}} |
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