三角形ABCにおいて、与えられた情報から残りの辺と角の大きさを求めます。 (1) $a=3$, $A=60^\circ$, $B=45^\circ$ (2) $a=2$, $b=\sqrt{6}$, $A=45^\circ$

1. 問題の内容

三角形ABCにおいて、与えられた情報から残りの辺と角の大きさを求めます。

(1) $a=3$, $A=60^\circ$, $B=45^\circ$

(2) $a=2$, $b=\sqrt{6}$, $A=45^\circ$

2. 解き方の手順

(1) $a=3$, $A=60^\circ$, $B=45^\circ$の場合

まず、角Cを求めます。三角形の内角の和は$180^\circ$なので、

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

正弦定理より、$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$が成り立ちます。

$\frac{3}{\sin 60^\circ} = \frac{b}{\sin 45^\circ}$

$b = \frac{3 \sin 45^\circ}{\sin 60^\circ} = \frac{3 \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{3}\sqrt{2} = \sqrt{6}$

$\frac{3}{\sin 60^\circ} = \frac{c}{\sin 75^\circ}$

$c = \frac{3 \sin 75^\circ}{\sin 60^\circ}$

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$c = \frac{3 (\frac{\sqrt{6}+\sqrt{2}}{4})}{\frac{\sqrt{3}}{2}} = \frac{3(\sqrt{6}+\sqrt{2})}{4} \cdot \frac{2}{\sqrt{3}} = \frac{3(\sqrt{6}+\sqrt{2})}{2\sqrt{3}} = \frac{\sqrt{3}(\sqrt{6}+\sqrt{2})}{2} = \frac{\sqrt{18}+\sqrt{6}}{2} = \frac{3\sqrt{2}+\sqrt{6}}{2}$

(2) $a=2$, $b=\sqrt{6}$, $A=45^\circ$の場合

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

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

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

$\sin B = \frac{\sqrt{3}}{2}$より、$B=60^\circ$または$B=120^\circ$

(i) $B=60^\circ$の場合

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

$c = \frac{a \sin C}{\sin A} = \frac{2 \sin 75^\circ}{\sin 45^\circ} = \frac{2 \cdot \frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{6}+\sqrt{2}}{2} \cdot \frac{2}{\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}} = \sqrt{3}+1$

(ii) $B=120^\circ$の場合

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

$c = \frac{a \sin C}{\sin A} = \frac{2 \sin 15^\circ}{\sin 45^\circ}$

$\sin 15^\circ = \sin (45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}$

$c = \frac{2 \cdot \frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{6}-\sqrt{2}}{2} \cdot \frac{2}{\sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}} = \sqrt{3}-1$

3. 最終的な答え

(1) $b=\sqrt{6}$, $c=\frac{3\sqrt{2}+\sqrt{6}}{2}$, $C=75^\circ$

(2) (i) $B=60^\circ$, $c=\sqrt{3}+1$, $C=75^\circ$

(ii) $B=120^\circ$, $c=\sqrt{3}-1$, $C=15^\circ$