三角形ABCにおいて、以下の条件で残りの辺の長さと角の大きさを求める問題です。 (1) $b = \sqrt{6}, c = \sqrt{3} - 1, A = 45^\circ$ のとき、$a, B, C$ (2) $a = 1 + \sqrt{3}, b = 2, c = \sqrt{6}$ のとき、$A, B, C$

1. 問題の内容

三角形ABCにおいて、以下の条件で残りの辺の長さと角の大きさを求める問題です。

(1) $b = \sqrt{6}, c = \sqrt{3} - 1, A = 45^\circ$ のとき、$a, B, C$

(2) $a = 1 + \sqrt{3}, b = 2, c = \sqrt{6}$ のとき、$A, B, C$

2. 解き方の手順

(1)

余弦定理より、$a^2 = b^2 + c^2 - 2bc \cos A$

$a^2 = (\sqrt{6})^2 + (\sqrt{3}-1)^2 - 2\sqrt{6}(\sqrt{3}-1) \cos 45^\circ$

$a^2 = 6 + (3 - 2\sqrt{3} + 1) - 2\sqrt{6}(\sqrt{3}-1) \frac{\sqrt{2}}{2}$

$a^2 = 10 - 2\sqrt{3} - \sqrt{3}( \sqrt{3}-1) \cdot 2$

$a^2 = 10 - 2\sqrt{3} - (3-\sqrt{3})\cdot2$

$a^2 = 10 - 2\sqrt{3} - 6 + 2\sqrt{3} = 4$

$a = 2$

正弦定理より、$\frac{a}{\sin A} = \frac{b}{\sin B}$

$\frac{2}{\sin 45^\circ} = \frac{\sqrt{6}}{\sin B}$

$\sin B = \frac{\sqrt{6} \sin 45^\circ}{2} = \frac{\sqrt{6} \frac{\sqrt{2}}{2}}{2} = \frac{\sqrt{12}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$

$B = 60^\circ$ または $B = 120^\circ$

$A+B+C = 180^\circ$なので、$B = 60^\circ$のとき$C = 180^\circ - 45^\circ - 60^\circ = 75^\circ$

$B = 120^\circ$のとき$C = 180^\circ - 45^\circ - 120^\circ = 15^\circ$

$C = 75^\circ$のとき、$\frac{c}{\sin C} = \frac{a}{\sin A}$より$\sin 75^\circ = \sin(45^\circ+30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$.

$\frac{\sqrt{3}-1}{(\sqrt{6}+\sqrt{2})/4} = \frac{4(\sqrt{3}-1)}{\sqrt{2}(\sqrt{3}+1)} = \frac{2\sqrt{2}(\sqrt{3}-1)}{\sqrt{3}+1} = \frac{2\sqrt{2} (\sqrt{3}-1)(\sqrt{3}-1)}{2} = \sqrt{2}(3-2\sqrt{3}+1) = \sqrt{2}(4-2\sqrt{3}) = 4\sqrt{2} - 2\sqrt{6}$. $\frac{a}{\sin A} = \frac{2}{\sqrt{2}/2} = 2\sqrt{2}$. この値と一致しないので、$C = 75^\circ$ は誤り。

$C = 15^\circ$のとき、

$\frac{c}{\sin C} = \frac{a}{\sin A}$

$\frac{\sqrt{3}-1}{\sin 15^\circ} = 2\sqrt{2}$

$\sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$\sin(45-30) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$なので、$C=15^\circ$が正しい。

よって$B = 120^\circ$。

(2)

余弦定理より、

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{2^2 + (\sqrt{6})^2 - (1+\sqrt{3})^2}{2 \cdot 2 \cdot \sqrt{6}} = \frac{4 + 6 - (1 + 2\sqrt{3} + 3)}{4\sqrt{6}} = \frac{10 - 4 - 2\sqrt{3}}{4\sqrt{6}} = \frac{6 - 2\sqrt{3}}{4\sqrt{6}} = \frac{3 - \sqrt{3}}{2\sqrt{6}} = \frac{3\sqrt{6} - 3\sqrt{2}}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$A = 75^\circ$

余弦定理より、

$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(1+\sqrt{3})^2 + (\sqrt{6})^2 - 2^2}{2(1+\sqrt{3})\sqrt{6}} = \frac{1 + 2\sqrt{3} + 3 + 6 - 4}{2\sqrt{6}(1+\sqrt{3})} = \frac{6 + 2\sqrt{3}}{2\sqrt{6}(1+\sqrt{3})} = \frac{3 + \sqrt{3}}{\sqrt{6}(1+\sqrt{3})} = \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{6}(1+\sqrt{3})} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$B = 45^\circ$

$C = 180^\circ - 75^\circ - 45^\circ = 60^\circ$

3. 最終的な答え

(1) $a = 2, B = 120^\circ, C = 15^\circ$

(2) $A = 75^\circ, B = 45^\circ, C = 60^\circ$