高専の数学解説

第14章 14.11

S=12ab⋅sin⁡CS = \frac{1}{2}ab \cdot \sin C

b=sin⁡Bsin⁡Aab = \frac{\sin B}{\sin A}a

sin⁡A=sin⁡(180−(B+C))=sin⁡180⋅cos⁡(B+C)−cos⁡180⋅sin⁡(B+C)=sin⁡(B+C)\sin A = \sin\bigl(180 - (B+C)\bigr) = \sin 180 \cdot \cos(B+C) - \cos 180 \cdot \sin(B+C) = \sin(B+C)

これから

b=sin⁡Bsin⁡(B+C)ab = \frac{\sin B}{\sin(B+C)}a

S=12a⋅sin⁡Bsin⁡(B+C)a⋅sin⁡C=a2sin⁡B⋅sin⁡C2sin⁡(B+C)S = \frac{1}{2}a \cdot \frac{\sin B}{\sin(B+C)}a \cdot \sin C = \frac{a^2 \sin B \cdot \sin C}{2\sin(B+C)}

解説: r31bn1z