r30 vs r31
......
5757
이때, [math(\sin^{2}{B}+\cos^{2}{B}=1)]이니 [math(\sin {B}= 1- \cos {B})] 를 얻어서 다음을 만든다.
5858
5959
[math(\displaystyle \begin{aligned} \triangle {\rm ABC}&=\dfrac{1}{2}ac\sin{B} =\dfrac{1}{2}ac \sqrt{1-\left(\frac{a^2+c^2-b^{2}}{2ac} \right)^{2}} \end{aligned} )]
60
61
[math(\displaystyle \begin{aligned} \triangle {\rm ABC}&= \dfrac{1}{4}\sqrt{4a^{2}c^{2}-(a^2+c^2-b^{2})^{2}} \end{aligned} )]