三道有关的求角度几何题。加加减减而已，但凑出来可不那么容易

mathematica

做且仅做第一题,其余依次类推

1. Clear["Global`*"];(*mathematica11.2,win7(64bit)Clear all variables*)
   
2. (*利用四边形角元塞瓦定理来解决问题,从A点的两个角开始,然后依次BCD三点*)
   
3. ans=NSolve[{
   
4.     Sin[50*Degree]/Sin[30*Degree]*
   
5.     Sin[60*Degree]/Sin[20*Degree]*
   
6.     Sin[70*Degree]/Sin[(90-x)*Degree]*
   
7.     Sin[x*Degree]/Sin[40*Degree]==1,
   
8.     0<x<180(*限制变量范围*)
   
9. },{x},100]
   

复制代码

求解结果
\[\{\{x\to 10.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\}\}\]

mathematica

王守恩 发表于 2021-2-4 14:11
重温 9 楼，解法不好(太狭窄)，奢望对整理思路有帮助。

第 1 题(看图：x 为起点，顺时针走一圈)。x=1 ...

看样子你似乎知道角元塞瓦定理！

creasson

再来解第一题：
\[D = A + \frac{{\left( {1 - {w^2}} \right){z^2}}}{{\left( {1 - {w^2}{z^2}} \right)}}\left( {B - A} \right),z = {e^{i\angle DAB}} = {e^{i\frac{{80}}{{180}}\pi }},w = {e^{i\angle DBA}} = {e^{i\frac{{60}}{{180}}\pi }}\]
\[E = A + \frac{{\left( {1 - {q^2}} \right){p^2}}}{{\left( {1 - {q^2}{p^2}} \right)}}\left( {B - A} \right),p = {e^{i\angle EAB}} = {e^{i\frac{{30}}{{180}}\pi }},q = {e^{i\angle EBA}} = {e^{i\frac{{80}}{{180}}\pi }}\]
\[{e^{i2x}} = \frac{{E - D}}{{B - D}}/conj\left( {\frac{{E - D}}{{B - D}}} \right) = \frac{{{w^2}( - {p^2} + {p^2}{q^2} + {z^2} - {p^2}{q^2}{z^2} - {w^2}{z^2} + {p^2}{w^2}{z^2})}}{{{q^2} - {p^2}{q^2} - {w^2} + {p^2}{q^2}{w^2} + {w^2}{z^2} - {q^2}{w^2}{z^2}}} = {e^{i\frac{\pi }{9}}}\]
\[x = \frac{\pi }{{18}}\]