dlpg070 发表于 2020-4-28 19:10
应约画了2个图形,供进一步分析参考
我们没负担，不妨自由一点。记P,Q交点为D，∠OAD=a，∠AOD=b，则
\(DO=\sin(a),DA=\sin(b),OA=\sin(a+b)\)
\(DB=k\sin(a),DC=k\sin(b),BC=k\sin(a+b)\)
\(OB=(k+1)\sin(a)，AC=(k+1)\sin(b)\)
1，\(\D S(梯形ABCO)=S(三角形ADO)+S(三角形形ODC)+S(三角形形CDB)+S(三角形形BDA)\)
\(=\frac{1}{2}\sin(b)\sin(a)\sin(a+b)+\frac{1}{2}\sin(a)(k\sin(b))\sin(a+b)\)
\(+\frac{1}{2}(k\sin(b))(k\sin(a))\sin(a+b)+\frac{1}{2}(k\sin(a))\sin(b)\sin(a+b)\)
\(=\frac{1}{2}\sin(a)\sin(b)\sin(a+b)(k+1)^2\)
2，\(\D S(梯形ABCO)=S(三角形OAC)+S(三角形形ACB)\)
\(=\frac{1}{2}\sin(a+b)((k+1)\sin(b))\sin(a)+\frac{1}{2}(k\sin(a+b))((k+1)\sin(b))\sin(a)\)
\(=\frac{1}{2}\sin(a)\sin(b)\sin(a+b)(k+1)^2\)
3，\(\D S(梯形ABCO)=S(三角形AOB)+S(三角形形OBC)\)
\(=\frac{1}{2}\sin(a+b)((k+1)\sin(a))\sin(b)+\frac{1}{2}((k+1)\sin(a))(k\sin(a+b))\sin(b)\)
\(=\frac{1}{2}\sin(a)\sin(b)\sin(a+b)(k+1)^2\)
4，\(\D S(梯形ABCO)=\frac{1}{2}OB*AC\sin(a+b)\)
\(=\frac{1}{2}((k+1)\sin(a))((k+1)\sin(b))\sin(a+b)\)
\(=\frac{1}{2}\sin(a)\sin(b)\sin(a+b)(k+1)^2\)
页:
1
[2]