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1章 数と式
2章 集合と論証
3章 2次関数
5章 データの分析
4章 図形と計量
Readiness check
p.122
問1
\({\small (1)}~4\sqrt{6}\) \({\small (2)}~{\large \frac{3\sqrt{7}}{4}}\)
問1
\({\small (1)}~4\sqrt{6}\) \({\small (2)}~{\large \frac{3\sqrt{7}}{4}}\)
p.122
問2
\({\small (1)}~{\large \frac{2\sqrt{7}}{7}}\) \({\small (2)}~{\large \frac{\sqrt{10}}{4}}\)
問2
\({\small (1)}~{\large \frac{2\sqrt{7}}{7}}\) \({\small (2)}~{\large \frac{\sqrt{10}}{4}}\)
p.122
問3
\({\large \frac{9}{2}}\)
問3
\({\large \frac{9}{2}}\)
p.123
問4
\({\small (1)}~x=5\) \({\small (2)}~x=2\)
問4
\({\small (1)}~x=5\) \({\small (2)}~x=2\)
p.123
問5
\({\small (1)}~x={\large \frac{\sqrt{2}}{2}}~,~y={\large \frac{\sqrt{2}}{2}}\)
\({\small (2)}~x={\large \frac{\sqrt{3}}{2}}~,~y={\large \frac{1}{2}}\)
問5
\({\small (1)}~x={\large \frac{\sqrt{2}}{2}}~,~y={\large \frac{\sqrt{2}}{2}}\)
\({\small (2)}~x={\large \frac{\sqrt{3}}{2}}~,~y={\large \frac{1}{2}}\)
1節 鋭角の三角比
p.125
問1
\({\small (1)}~\)
\(\sin{A}={\large \frac{2}{3}}~,~\cos{A}={\large \frac{\sqrt{5}}{3}}\)
\(\tan{A}={\large \frac{2}{\sqrt{5}}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{3}{\sqrt{10}}}~,~\cos{A}={\large \frac{1}{\sqrt{10}}}\)
\(\tan{A}=3\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{5}{13}}~,~\cos{A}={\large \frac{12}{13}}\)
\(\tan{A}={\large \frac{5}{12}}\)
問1
\({\small (1)}~\)
\(\sin{A}={\large \frac{2}{3}}~,~\cos{A}={\large \frac{\sqrt{5}}{3}}\)
\(\tan{A}={\large \frac{2}{\sqrt{5}}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{3}{\sqrt{10}}}~,~\cos{A}={\large \frac{1}{\sqrt{10}}}\)
\(\tan{A}=3\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{5}{13}}~,~\cos{A}={\large \frac{12}{13}}\)
\(\tan{A}={\large \frac{5}{12}}\)
p.126
問2
\({\small (1)}~\)
\(\sin{A}={\large \frac{3}{\sqrt{13}}}~,~\cos{A}={\large \frac{2}{\sqrt{13}}}\)
\(\tan{A}={\large \frac{3}{2}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{\sqrt{6}}{3}}~,~\cos{A}={\large \frac{\sqrt{3}}{3}}\)
\(\tan{A}=\sqrt{2}\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{3}{4}}~,~\cos{A}={\large \frac{\sqrt{7}}{4}}\)
\(\tan{A}={\large \frac{3}{\sqrt{7}}}\)
→ 直角三角形と三角比
問2
\({\small (1)}~\)
\(\sin{A}={\large \frac{3}{\sqrt{13}}}~,~\cos{A}={\large \frac{2}{\sqrt{13}}}\)
\(\tan{A}={\large \frac{3}{2}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{\sqrt{6}}{3}}~,~\cos{A}={\large \frac{\sqrt{3}}{3}}\)
\(\tan{A}=\sqrt{2}\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{3}{4}}~,~\cos{A}={\large \frac{\sqrt{7}}{4}}\)
\(\tan{A}={\large \frac{3}{\sqrt{7}}}\)
→ 直角三角形と三角比
p.127
問3
\(\sin{69^\circ}=0.9336\)
\(\tan{69^\circ}=2.6051\)
問3
\(\sin{69^\circ}=0.9336\)
\(\tan{69^\circ}=2.6051\)
p.127
問4
\({\small (1)}~0.6428\)
\({\small (2)}~0.5299\)
\({\small (3)}~0.4663\)
問4
\({\small (1)}~0.6428\)
\({\small (2)}~0.5299\)
\({\small (3)}~0.4663\)
p.127
問5
\({\small (1)}~75^\circ\) \({\small (2)}~23^\circ\) \({\small (3)}~10^\circ\)
問5
\({\small (1)}~75^\circ\) \({\small (2)}~23^\circ\) \({\small (3)}~10^\circ\)
p.128
問6
\({\rm AC}=4\sqrt{3}~,~{\rm BC}=4\)
問6
\({\rm AC}=4\sqrt{3}~,~{\rm BC}=4\)
p.128
問7
\({\rm BC}=2\)
問7
\({\rm BC}=2\)
p.129
問8
\(869\) m
問8
\(869\) m
p.129
問9
\({\small (1)}~58^\circ\) \({\small (2)}~42^\circ\)
問9
\({\small (1)}~58^\circ\) \({\small (2)}~42^\circ\)
p.131
問10
\(\cos{A}={\large \frac{\sqrt{15}}{4}}~,~\tan{A}={\large \frac{1}{\sqrt{15}}}\)
→ 三角比の相互関係の公式(鋭角)
問10
\(\cos{A}={\large \frac{\sqrt{15}}{4}}~,~\tan{A}={\large \frac{1}{\sqrt{15}}}\)
→ 三角比の相互関係の公式(鋭角)
p.132
問11
\(\cos{A}={\large \frac{1}{\sqrt{10}}}~,~\sin{A}={\large \frac{3}{\sqrt{10}}}\)
→ 三角比の相互関係の公式(鋭角)
問11
\(\cos{A}={\large \frac{1}{\sqrt{10}}}~,~\sin{A}={\large \frac{3}{\sqrt{10}}}\)
→ 三角比の相互関係の公式(鋭角)
p.133
問12
\({\small (1)}~\cos{3^\circ}\) \({\small (2)}~\sin{34^\circ}\)
\({\small (3)}~{\large \frac{1}{\tan{18^\circ}}}\)
→ 余角の公式
問12
\({\small (1)}~\cos{3^\circ}\) \({\small (2)}~\sin{34^\circ}\)
\({\small (3)}~{\large \frac{1}{\tan{18^\circ}}}\)
→ 余角の公式
Training
p.134
1
\({\small (1)}~\)
\(\sin{A}={\large \frac{2\sqrt{6}}{7}}~,~\cos{A}={\large \frac{5}{7}}\)
\(\tan{A}={\large \frac{2\sqrt{6}}{5}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{5}{13}}~,~\cos{A}={\large \frac{12}{13}}\)
\(\tan{A}={\large \frac{5}{12}}\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{\sqrt{15}}{5}}~,~\cos{A}={\large \frac{\sqrt{10}}{5}}\)
\(\tan{A}={\large \frac{\sqrt{6}}{2}}\)
1
\({\small (1)}~\)
\(\sin{A}={\large \frac{2\sqrt{6}}{7}}~,~\cos{A}={\large \frac{5}{7}}\)
\(\tan{A}={\large \frac{2\sqrt{6}}{5}}\)
\({\small (2)}~\)
\(\sin{A}={\large \frac{5}{13}}~,~\cos{A}={\large \frac{12}{13}}\)
\(\tan{A}={\large \frac{5}{12}}\)
\({\small (3)}~\)
\(\sin{A}={\large \frac{\sqrt{15}}{5}}~,~\cos{A}={\large \frac{\sqrt{10}}{5}}\)
\(\tan{A}={\large \frac{\sqrt{6}}{2}}\)
p.134
2
\({\small (1)}~{\rm AC}={\large \frac{5\sqrt{2}}{2}}~,~{\rm BC}={\large \frac{5\sqrt{2}}{2}}\)
\({\small (2)}~{\rm AC}=4\sqrt{3}~,~{\rm BC}=8\)
2
\({\small (1)}~{\rm AC}={\large \frac{5\sqrt{2}}{2}}~,~{\rm BC}={\large \frac{5\sqrt{2}}{2}}\)
\({\small (2)}~{\rm AC}=4\sqrt{3}~,~{\rm BC}=8\)
p.134
3
\(5.1\) m
3
\(5.1\) m
p.134
4
\(\cos{A}={\large \frac{3\sqrt{10}}{10}}~,~\tan{A}={\large \frac{1}{3}}\)
4
\(\cos{A}={\large \frac{3\sqrt{10}}{10}}~,~\tan{A}={\large \frac{1}{3}}\)
p.134
5
\(\sin{A}={\large \frac{\sqrt{55}}{8}}~,~\tan{A}={\large \frac{\sqrt{55}}{3}}\)
5
\(\sin{A}={\large \frac{\sqrt{55}}{8}}~,~\tan{A}={\large \frac{\sqrt{55}}{3}}\)
p.134
6
\(\cos{A}={\large \frac{\sqrt{6}}{3}}~,~\sin{A}={\large \frac{\sqrt{3}}{3}}\)
6
\(\cos{A}={\large \frac{\sqrt{6}}{3}}~,~\sin{A}={\large \frac{\sqrt{3}}{3}}\)
p.134
7
\({\small (1)}~\cos{28^\circ}\) \({\small (2)}~\sin{12^\circ}\)
\({\small (3)}~{\large \frac{1}{\tan{10^\circ}}}\)
7
\({\small (1)}~\cos{28^\circ}\) \({\small (2)}~\sin{12^\circ}\)
\({\small (3)}~{\large \frac{1}{\tan{10^\circ}}}\)
2節 三角比の拡張
p.136
問1
\({\small (1)}~\sin{135^\circ}={\large \frac{1}{\sqrt{2}}}~,~\cos{135^\circ}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{135^\circ}=-1\)
\({\small (2)}~\sin{150^\circ}={\large \frac{1}{2}}~,~\cos{150^\circ}=-{\large \frac{\sqrt{3}}{2}}\)
\(\tan{135^\circ}=-{\large \frac{1}{\sqrt{3}}}\)
→ 三角比の拡張
問1
\({\small (1)}~\sin{135^\circ}={\large \frac{1}{\sqrt{2}}}~,~\cos{135^\circ}=-{\large \frac{1}{\sqrt{2}}}\)
\(\tan{135^\circ}=-1\)
\({\small (2)}~\sin{150^\circ}={\large \frac{1}{2}}~,~\cos{150^\circ}=-{\large \frac{\sqrt{3}}{2}}\)
\(\tan{135^\circ}=-{\large \frac{1}{\sqrt{3}}}\)
→ 三角比の拡張
p.138
問2
\({\small (1)}~\theta=60^\circ~,~120^\circ\)
\({\small (2)}~\theta=90^\circ\)
\({\small (3)}~\theta=60^\circ\)
\({\small (4)}~\theta=150^\circ\)
→ 三角比と方程式
問2
\({\small (1)}~\theta=60^\circ~,~120^\circ\)
\({\small (2)}~\theta=90^\circ\)
\({\small (3)}~\theta=60^\circ\)
\({\small (4)}~\theta=150^\circ\)
→ 三角比と方程式
p.141
問4
\({\small (1)}~\cos{\theta}=-{\large \frac{\sqrt{5}}{3}}~,~\tan{\theta}=-{\large \frac{2}{\sqrt{5}}}\)
\({\small (2)}~\sin{\theta}={\large \frac{3}{5}}~,~\tan{\theta}=-{\large \frac{3}{4}}\)
→ 三角比の相互関係の公式(鈍角)
問4
\({\small (1)}~\cos{\theta}=-{\large \frac{\sqrt{5}}{3}}~,~\tan{\theta}=-{\large \frac{2}{\sqrt{5}}}\)
\({\small (2)}~\sin{\theta}={\large \frac{3}{5}}~,~\tan{\theta}=-{\large \frac{3}{4}}\)
→ 三角比の相互関係の公式(鈍角)
p.141
問5
\(\cos{\theta}=-{\large \frac{3}{\sqrt{10}}}~,~\sin{\theta}={\large \frac{1}{\sqrt{10}}}\)
→ 三角比の相互関係の公式(鈍角)
問5
\(\cos{\theta}=-{\large \frac{3}{\sqrt{10}}}~,~\sin{\theta}={\large \frac{1}{\sqrt{10}}}\)
→ 三角比の相互関係の公式(鈍角)
Training
p.143
8
\({\small (1)}~\theta=45^\circ~,~135^\circ\)
\({\small (2)}~\theta=120^\circ\)
\({\small (3)}~\theta=135^\circ\)
8
\({\small (1)}~\theta=45^\circ~,~135^\circ\)
\({\small (2)}~\theta=120^\circ\)
\({\small (3)}~\theta=135^\circ\)
p.143
9
\({\small (1)}~\sin{\theta}={\large \frac{\sqrt{15}}{4}}~,~\tan{\theta}=-\sqrt{15}\)
\({\small (2)}~\cos{\theta}={\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}={\large \frac{\sqrt{2}}{2}}\)
または、
\(\cos{\theta}=-{\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}=-{\large \frac{\sqrt{2}}{2}}\)
\({\small (3)}~\sin{\theta}={\large \frac{2}{3}}~,~\cos{\theta}={\large \frac{\sqrt{5}}{3}}\)
9
\({\small (1)}~\sin{\theta}={\large \frac{\sqrt{15}}{4}}~,~\tan{\theta}=-\sqrt{15}\)
\({\small (2)}~\cos{\theta}={\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}={\large \frac{\sqrt{2}}{2}}\)
または、
\(\cos{\theta}=-{\large \frac{\sqrt{6}}{3}}~,~\tan{\theta}=-{\large \frac{\sqrt{2}}{2}}\)
\({\small (3)}~\sin{\theta}={\large \frac{2}{3}}~,~\cos{\theta}={\large \frac{\sqrt{5}}{3}}\)
p.143
10
\({\small (1)}~\cos{20^\circ}\) \({\small (2)}~-\sin{5^\circ}\)
\({\small (3)}~-{\large \frac{1}{\tan{40^\circ}}}\)
10
\({\small (1)}~\cos{20^\circ}\) \({\small (2)}~-\sin{5^\circ}\)
\({\small (3)}~-{\large \frac{1}{\tan{40^\circ}}}\)
3節 三角形への応用
p.148
問4
\(c=2\sqrt{2}~,~A=30^\circ~,~C=105^\circ\)
問4
\(c=2\sqrt{2}~,~A=30^\circ~,~C=105^\circ\)
p.150
問6
\({\small (1)}~-{\large \frac{3}{7}}\) \({\small (2)}~{\large \frac{2\sqrt{10}}{7}}\) \({\small (3)}~6\sqrt{10}\)
→ 三角形の面積(三角比)
問6
\({\small (1)}~-{\large \frac{3}{7}}\) \({\small (2)}~{\large \frac{2\sqrt{10}}{7}}\) \({\small (3)}~6\sqrt{10}\)
→ 三角形の面積(三角比)
p.154
問8
\(141\) m
問8
\(141\) m
p.156
発展1
\(10\sqrt{2}\)
発展1
\(10\sqrt{2}\)
Training
p.157
11
\({\small (1)}~b=5\sqrt{2}~,~R=5\)
\({\small (2)}~b=\sqrt{3}~,~A=45^\circ\)
11
\({\small (1)}~b=5\sqrt{2}~,~R=5\)
\({\small (2)}~b=\sqrt{3}~,~A=45^\circ\)
p.157
12
\({\small (1)}~\sqrt{7}\) \({\small (2)}~13\)
12
\({\small (1)}~\sqrt{7}\) \({\small (2)}~13\)
p.157
13
\({\small (1)}~45^\circ\) \({\small (2)}~135^\circ\)
13
\({\small (1)}~45^\circ\) \({\small (2)}~135^\circ\)
p.157
14
\(c=\sqrt{6}~,~A=45^\circ~,~B=15^\circ\)
14
\(c=\sqrt{6}~,~A=45^\circ~,~B=15^\circ\)
p.157
15
\({\small (1)}~6\sqrt{3}\) \({\small (2)}~12\sqrt{5}\)
15
\({\small (1)}~6\sqrt{3}\) \({\small (2)}~12\sqrt{5}\)
p.157
16
\({\small (1)}~\sqrt{7}\) \({\small (2)}~60^\circ\) \({\small (3)}~2\sqrt{3}\)
16
\({\small (1)}~\sqrt{7}\) \({\small (2)}~60^\circ\) \({\small (3)}~2\sqrt{3}\)
p.157
17
\({\large \frac{1}{3}}\)
17
\({\large \frac{1}{3}}\)
Level Up 図形と計量
p.158
1
\(5+5\sqrt{3}\)
1
\(5+5\sqrt{3}\)
p.158
2
\({\small (1)}~0\) \({\small (2)}~-2\) \({\small (3)}~2\)
2
\({\small (1)}~0\) \({\small (2)}~-2\) \({\small (3)}~2\)
p.158
3
\(1\) または \(2\)
3
\(1\) または \(2\)
p.158
4
\(120^\circ\)
4
\(120^\circ\)
p.158
5
[証明] 正弦定理より、
\({\large \frac{c}{\sin{{\rm C}}}}=2R\)
これより、
\(\sin{{\rm C}}={\large \frac{c}{2R}}\)
また、面積公式は、
\(S={\large \frac{1}{2}}bc\sin{{\rm C}}\)
上の式を代入すると、
\(S={\large \frac{1}{2}}bc{\large \frac{c}{2R}}\)
したがって、
\(S={\large \frac{abc}{4R}}\) [終]
5
[証明] 正弦定理より、
\({\large \frac{c}{\sin{{\rm C}}}}=2R\)
これより、
\(\sin{{\rm C}}={\large \frac{c}{2R}}\)
また、面積公式は、
\(S={\large \frac{1}{2}}bc\sin{{\rm C}}\)
上の式を代入すると、
\(S={\large \frac{1}{2}}bc{\large \frac{c}{2R}}\)
したがって、
\(S={\large \frac{abc}{4R}}\) [終]
p.158
6
\({\small (1)}~\triangle {\rm ABC}=x~,~\triangle {\rm ACD}={\large \frac{3}{4}}x\)
\({\small (2)}~x={\large \frac{12\sqrt{3}}{7}}\)
→ 角の二等分線の長さ
6
\({\small (1)}~\triangle {\rm ABC}=x~,~\triangle {\rm ACD}={\large \frac{3}{4}}x\)
\({\small (2)}~x={\large \frac{12\sqrt{3}}{7}}\)
→ 角の二等分線の長さ
p.159
7
\({\small (1)}~{\rm BD}=\sqrt{37}~,~\cos{\angle{\rm BAD}}=-{\large \frac{1}{10}}\)
\({\small (2)}~{\large \frac{21\sqrt{11}}{4}}\)
7
\({\small (1)}~{\rm BD}=\sqrt{37}~,~\cos{\angle{\rm BAD}}=-{\large \frac{1}{10}}\)
\({\small (2)}~{\large \frac{21\sqrt{11}}{4}}\)
p.159
8
\({\small (1)}~{\large \frac{\sqrt{2}}{2}}\) \({\small (2)}~60^\circ\)
\({\small (3)}~{\large \frac{3\sqrt{3}}{2}}\) \({\small (4)}~{\large \frac{\sqrt{6}}{3}}\)
8
\({\small (1)}~{\large \frac{\sqrt{2}}{2}}\) \({\small (2)}~60^\circ\)
\({\small (3)}~{\large \frac{3\sqrt{3}}{2}}\) \({\small (4)}~{\large \frac{\sqrt{6}}{3}}\)
p.159
9
\(408\) m
9
\(408\) m
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