問題2の(1)と(2)について、$a$と$b$の値をそれぞれ求める。

1. 問題の内容

問題2の(1)と(2)について、$a$と$b$の値をそれぞれ求める。

2. 解き方の手順

(1)

$\angle C = 180^\circ - 45^\circ - 30^\circ = 105^\circ$

正弦定理より、

$\frac{a}{\sin{45^\circ}} = \frac{1}{\sin{30^\circ}}$

$a = \frac{\sin{45^\circ}}{\sin{30^\circ}} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$

$\frac{b}{\sin{105^\circ}} = \frac{1}{\sin{30^\circ}}$

$\sin{105^\circ} = \sin{(60^\circ+45^\circ)} = \sin{60^\circ}\cos{45^\circ} + \cos{60^\circ}\sin{45^\circ} = \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$

$b = \frac{\sin{105^\circ}}{\sin{30^\circ}} = \frac{\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{1}{2}} = \frac{\sqrt{6}+\sqrt{2}}{2}$

(2)

余弦定理より

$a^2 = b^2 + 2^2 - 2\cdot b\cdot 2 \cos{60^\circ}$

$a^2 = b^2 + 4 - 4b\cdot \frac{1}{2}$

$a^2 = b^2 - 2b + 4$

正弦定理より

$\frac{2}{\sin{60^\circ}} = \frac{b}{\sin{45^\circ}}$

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

また、

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

$a = \frac{2 \sin{45^\circ} \sin{75^\circ}}{\sin{60^\circ} \sin{45^\circ}}$

$\frac{a}{\sin{60^\circ}} = \frac{2}{\sin{60^\circ}}$より、$a=2$

$a^2 = b^2 - 2b + 4$に代入

$4 = b^2 - 2b + 4$

$b^2 - 2b = 0$

$b(b-2) = 0$

$b = 0$または$b=2$

$b=0$は不適

$b=2$

3. 最終的な答え

(1) $a=\sqrt{2}$, $b=\frac{\sqrt{6}+\sqrt{2}}{2}$

(2) $a=2$, $b = \frac{2\sqrt{6}}{3}$

あるいは

(2) $a=2$, $b=2$