|
|

楼主 |
发表于 2026-6-11 07:56
|
显示全部楼层
\(△ABC,\sin A\sin B=\sin^2C,\sin C=\frac{\sqrt{7}}{4}=△ABC面积,求△ABC周长\)
\(由\ \sin A\sin B=\sin^2C,=>a*b=c^2,由\ \sin C=\frac{\sqrt{7}}{4},=>\cos C=\frac{3}{4}\)
\(△ABC面积=\frac{ab\sin C}{2}=\frac{\sqrt{7}}{4},=>ab=2,c=\sqrt{2}\)
\((\sqrt{2})^2=a^2+b^2-2ab\cos C=a^2+b^2-2*2*\frac{3}{4},=>a^2+b^2=5\)
\((a+b)^2=a^2+b^2+2ab=5+2*2,=>a+b=3\)
\(△ABC周长=a+b+c=3+\sqrt{2}。注意三边=1,2,\sqrt{2}是唯一的。\) |
|