【高校数学Ⅰ】3-3-1 正弦定理｜問題集

正弦定理より、
\(\displaystyle \frac{3}{\sin30^{\circ}}=\frac{a}{\sin45^{\circ}}\)
\(\displaystyle \frac{1}{2}a=\frac{1}{\sqrt{2}}\times3\)
\(\displaystyle a=\frac{6}{\sqrt{2}}\)
\(a=3\sqrt{2}\)

正弦定理より、
\(\displaystyle \frac{2}{\sin120^{\circ}}=2R\)
\(\displaystyle 2\div\frac{\sqrt{3}}{2}=2R\)
\(\displaystyle R=\frac{2}{\sqrt{3}}\)
\(\displaystyle R=\frac{2\sqrt{3}}{3}\)

正弦定理より、
\(\displaystyle \frac{\sqrt{3}}{\sin B}=\frac{3}{\sin60^{\circ}}\)
\(\displaystyle 3\sin B=\sqrt{3}\times\frac{\sqrt{3}}{2}\)
\(\displaystyle \sin B=\frac{1}{2}\)
\(B=30^{\circ},150^{\circ}\)
\(150^{\circ}\)は不適なため、
\(B=30^{\circ}\)

正弦定理より、
\(\displaystyle \frac{\sqrt{6}}{\sin C}=2\sqrt{3}\)
\(\displaystyle \sin C=\frac{\sqrt{6}}{2\sqrt{3}}\)
\(\displaystyle \sin C=\frac{1}{\sqrt{2}}\)
\(C=45^{\circ},135^{\circ}\)

正弦定理より、
\(\displaystyle \frac{2}{\sin60^{\circ}}=\frac{b}{\sin45^{\circ}}\)
\(\displaystyle \frac{\sqrt{3}}{2}b=2\times\frac{1}{\sqrt{2}}\)
\(\displaystyle b=\frac{2\sqrt{6}}{3}\)

正弦定理より、
\(\displaystyle \frac{6}{\sin120^{\circ}}=2R\)
\(\displaystyle 6\div\frac{\sqrt{3}}{2}=2R\)
\(\displaystyle R=\frac{6}{\sqrt{3}}\)
\(R=2\sqrt{3}\)