【数学I】4章：図形と計量（節末問題・章末問題）

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% 節末問題I4.1.1：（One More）★★★
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右の図のような直角三角形ABCを用いて，次の問いに答えよ．

(1) 辺ABの長さを求めよ．

(2) $\sin 22.5^{\circ},\cos 22.5^{\circ},\tan 22.5^{\circ}$の値を求めよ．

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% 節末問題I4.1.1の解答（One More）★★★
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(1) $\angle\mathrm{CBD}=180^\circ-(90^\circ+45^{\circ})=45^\circ$より，$\mathrm{CD=1,\mathrm{BD}=\sqrt{2}}$

また，$\angle\mathrm{ABC}=90^{\circ}-22.5^{\circ}=67.5^{\circ}$より，$\angle\mathrm{ABD}=67.5^{\circ}-45^{\circ}=22.5^{\circ}$

したがって，$\triangle\mathrm{ABD}$は$\mathrm{AD}=\mathrm{BD}$の二等辺三角形である．

よって，$\mathrm{CD}=1$より，$\mathrm{AC}=\mathrm{AD}+\mathrm{CD}=\sqrt{2}+1\cdots(\mathrm{i})$

$\triangle\mathrm{ABC}$において，三平方の定理より，

\[
\mathrm{AB}^2=(\sqrt{2}+1)^2+1^2=4+2\sqrt{2}
\]

$\mathrm{AB}>0$より，

\[
\mathrm{AB}=\sqrt{(\sqrt{2}+1)^2+1^2}=\sqrt{4+2\sqrt{2}}\cdots(\mathrm{ii})
\]

(2) $\triangle\mathrm{ABC}$において，(i)，(ii)，$\mathrm{BC}=1$より，

\begin{align*}
&\begin{aligned}\sin 22.5^{\circ}&=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{4+2\sqrt{2}}}=\sqrt{\frac{4-2\sqrt{2}}{(4+2\sqrt{2})(4-2\sqrt{2})}}\\
&=\sqrt{\frac{4-2\sqrt{2}}{8}}=\frac{\sqrt{2-\sqrt{2}}}{2}\end{aligned}\\
&\begin{aligned}\cos 22.5^{\circ}&=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{2}+1}{\sqrt{4+2\sqrt{2}}}=(\sqrt{2}+1)\sqrt{\frac{4-2\sqrt{2}}{(4+2\sqrt{2})(4-2\sqrt{2})}}\\
&=(\sqrt{2}+1)\sqrt{\frac{4-2\sqrt{2}}{8}}=\frac{(\sqrt{2}+1)\sqrt{2-\sqrt{2}}}{2}\\
&=\frac{\sqrt{(\sqrt{2}+1)^2(2-\sqrt{2})}}{2}=\frac{\sqrt{2+\sqrt{2}}}{2}\end{aligned}\\
&\tan 22.5^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}-1
\end{align*}

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% 節末問題I4.1.2：（One More）★★★
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鋭角$\theta$が$\tan\theta=\frac{5}{12}$を満たすとき，$\frac{\sin\theta}{1+\cos\theta}+\frac{\sin\theta}{1-\cos\theta}$の値を求めよ．

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% 節末問題I4.1.2の解答（One More）★★★
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\begin{align*}
\frac{\sin\theta}{1+\cos\theta}+\frac{\sin\theta}{1-\cos\theta}&=\frac{\sin\theta(1-\cos\theta)+\sin\theta(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}\\
&=\frac{\sin\theta-\sin\theta\cos\theta+\sin\theta+\sin\theta\cos\theta}{1-\cos^2\theta}\\
&=\frac{2\sin\theta}{\sin^2\theta}=\frac{2}{\sin\theta}
\end{align*}

また，

\[
\cos^2\theta=\frac{1}{1+\tan^2\theta}=\frac{1}{1+(\frac{5}{12})^2}=\frac{144}{169},
\]

\[
\sin^2\theta=1-\cos^2\theta=1-\frac{144}{169}=\frac{25}{169}
\]

$\theta$は鋭角であるから，$\sin\theta>0$

したがって，$\sin\theta=\frac{5}{13}$

よって，

\[
\frac{\sin\theta}{1+\cos\theta}+\frac{\sin\theta}{1-\cos\theta}=\frac{2}{\sin\theta}=2\div\frac{5}{13}=\frac{26}{5}
\]

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% 節末問題I4.1.2の別解
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$\tan\theta=\frac{5}{12}$より，

\[
\cos^2\theta=\frac{1}{1+\tan^2\theta}=\frac{1}{1+(\frac{5}{12})^2}=\frac{144}{169}
\]

$\theta$は鋭角であるから，$\cos\theta>0$

したがって，$\cos\theta=\frac{12}{13}$

また，$\sin\theta=\cos\theta\tan\theta=\frac{12}{13}\cdot\frac{5}{12}=\frac{5}{13}$

よって，これらを代入して，

\[
\frac{\sin\theta}{1+\cos\theta}+\frac{\sin\theta}{1-\cos\theta}=\frac{5}{13}\div\frac{25}{13}+\frac{5}{13}\div\frac{1}{13}=\frac{26}{5}
\]

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% 節末問題I4.1.3：（One More）★★
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$0^{\circ}\leqq\theta\leqq 180^{\circ}$において，$\frac{1}{\sqrt{3}}\sin\theta+\cos\theta=0$を満たす$\theta$の値を求めよ．

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% 節末問題I4.1.3の解答（One More）★★
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$\cos\theta=0$のとき，$\theta=90^{\circ}$であり，これは方程式を満たさない．

したがって，$\cos\theta\neq 0$

$\frac{1}{\sqrt{3}}\sin\theta+\cos\theta=0$より，$\sin\theta=-\sqrt{3}\cos\theta$

両辺を$\cos\theta$で割ると，$\frac{\sin\theta}{\cos\theta}=-{\sqrt{3}}$

よって，$\tan\theta=-{\sqrt{3}}$

$0^{\circ}\leqq\theta\leqq 180^{\circ}$であるから，$\theta=120^{\circ}$

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% 節末問題I4.1.4：（One More）★★★
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$\tan\theta+\frac{1}{\tan\theta}=3$のとき，次の式の値を求めよ．ただし，$0^{\circ}<\theta<45^{\circ}$とする．

(1) $\sin\theta\cos\theta$

(2) $\sin\theta+\cos\theta$

(3) $\sin\theta-\cos\theta$

(4) $\sin^3\theta-\cos^3\theta$

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(1)

\[
\tan\theta+\frac{1}{\tan\theta}=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}
\]

$\tan\theta+\frac{1}{\tan\theta}=3$であるから，$\frac{1}{\sin\theta\cos\theta}=3$

よって，$\sin\theta\cos\theta=\frac{1}{3}\cdots(\mathrm{i})$

(2)

\[
(\sin\theta+\cos\theta)^2=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=1+2\sin\theta\cos\theta
\]

(i)より，$(\sin\theta+\cos\theta)^2=1+2\cdot\frac{1}{3}=\frac{5}{3}$

$0^{\circ}<\theta<45^{\circ}$のとき，$\sin\theta>0,\cos\theta>0$より，$\sin\theta+\cos\theta>0$

よって，$\sin\theta+\cos\theta=\sqrt{\frac{5}{3}}=\frac{\sqrt{15}}{3}$

(3)

\[
(\sin\theta-\cos\theta)^2=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta=1-2\sin\theta\cos\theta
\]

(i)より，$(\sin\theta-\cos\theta)^2=1-2\cdot\frac{1}{3}=\frac{1}{3}$

$0^{\circ}<\theta<45^{\circ}$のとき，$\sin\theta<\cos\theta$より，$\sin\theta-\cos\theta<0$

よって，$\sin\theta-\cos\theta=-\sqrt{\frac{1}{3}}=-\frac{\sqrt{3}}{3}$

(4)

\begin{align*}
\sin^3\theta-\cos^3\theta&=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)\\
&=-\frac{\sqrt{3}}{3}\cdot(1+\frac{1}{3})=-\frac{4\sqrt{3}}{9}
\end{align*}

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% 節末問題I4.1.5：（One More）★★
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2直線$\sqrt{3}x+y=0\cdots(\mathrm{i}),ax+by+2=0\cdots(\mathrm{ii})$がある．直線(ii)は点$(-1,\sqrt{3})$を通り，2直線(i)，(ii)のなす角は$30^{\circ}$である．このとき，定数$a,b$の値を求めよ．

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% 節末問題I4.1.5の解答（One More）★★
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$\sqrt{3}x+y=0$より，$y=-\sqrt{3}x$

直線(i)が$x$軸の正の向きとのなす角を$\theta$とすると，$\tan\theta=-\sqrt{3}$

$0^{\circ}\leqq\theta<180^{\circ}$より，$\theta=120^{\circ}$

2直線(i)，(ii)のなす角が$30^{\circ}$であるから，直線(ii)が$x$軸の正の向きとのなす角は$90^{\circ}$または$150^{\circ}$

（ア）$90^{\circ}$のとき

直線(ii)は点$(-1,\sqrt{3})$を通り，$y$軸に平行であるから，$x=-1$

式変形すると，$2x+2=0$

したがって，これと(ii)を比較すると，$a=2,b=0$

（イ）$150^{\circ}$のとき

$\tan 150^{\circ}=-\frac{1}{\sqrt{3}}$より，直線(ii)の傾きは$-\frac{1}{\sqrt{3}}$

直線(ii)は点$(-1,\sqrt{3})$を通るから，$y=-\frac{1}{\sqrt{3}}(x+1)+\sqrt{3}$

式変形すると，$-x-\sqrt{3}y+2=0$

したがって，これと(ii)を比較すると，$a=-1,b=-\sqrt{3}$

よって，（ア），（イ）より，$a=2,b=0$または$a=-1,b=-\sqrt{3}$

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% 節末問題I4.2.1：（One More）★★
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$\triangle\mathrm{ABC}$の頂点$\mathrm{A},\mathrm{B},\mathrm{C}$の対辺をそれぞれ$a,b,c$とする．ただし，$a\neq b$とする．$C=60^\circ$のとき，$\frac{b}{c^2-a^2}+\frac{a}{c^2-b^2}$の値を求めよ．

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% 節末問題I4.2.1の解答（One More）★★
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余弦定理より，$c^2=a^2+b^2-2ab\cos 60^{\circ}=a^2+b^2-2ab\cdot\frac{1}{2}=a^2+b^2-ab$

整理すると，$c^2-a^2=b^2-ab$

また，$c^2-b^2=a^2-ab$

よって，これらを与えられた式に代入すると，

\begin{align*}
\frac{b}{c^2-a^2}+\frac{a}{c^2-b^2}&=\frac{b}{b^2-ab}+\frac{a}{a^2-ab}\\
&=\frac{b}{b(b-a)}+\frac{a}{a(a-b)}\\
&=\frac{1}{b-a}+\frac{1}{a-b}=0
\end{align*}

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% 節末問題I4.2.2：（One More）★★★
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$\triangle\mathrm{ABC}$において，$b=2,c=\sqrt{6},B=45^{\circ}$のとき，残りの辺の長さと角の大きさを求めよ．

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% 節末問題I4.2.2の解答（One More）★★★
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正弦定理より，$\frac{2}{\sin 45^{\circ}}=\frac{\sqrt{6}}{\sin C}$

したがって，$\sin C=\sqrt{6}\cdot\frac{\sin 45^{\circ}}{2}=\frac{\sqrt{3}}{2}$

$B=45^{\circ}$より，$0^{\circ}<C<135^{\circ}$であるから，

\[
C=60^{\circ},120^{\circ}
\]

(i) $C=60^{\circ}$のとき

\[
A=180^{\circ}-(B+C)=75^{\circ}
\]

余弦定理より，$2^2=a^2+(\sqrt{6})^2-2\cdot a\cdot\sqrt{6}\cdot\cos 45^{\circ}$

したがって，$a^2-2\sqrt{3}a+2=0$より，$a=\sqrt{3}\pm 1$

ここで，$A$が最大の角であるから，対辺$a$が最大の辺である．

ゆえに，$a>\sqrt{6}$より，$a=\sqrt{3}+1$

(ii) $C=120^{\circ}$のとき

\[
A=180^{\circ}-(B+C)=15^{\circ}
\]

余弦定理より，$2^2=a^2+(\sqrt{6})^2-2\cdot a\cdot\sqrt{6}\cdot\cos 45^{\circ}$

したがって，$a^2-2\sqrt{3}a+2=0$より，$a=\sqrt{3}\pm 1$

ここで，$A$が最小の角であるから，対辺$a$が最小の辺である．

ゆえに，$a<2$より，$a=\sqrt{3}-1$

よって，(i)，(ii)より，

\[
A=75^{\circ},C=60^{\circ},a=\sqrt{3}+1\text{または}A=15^{\circ},C=120^{\circ},a=\sqrt{3}-1
\]

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% 節末問題I4.2.3：（One More）★★★
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$\triangle\mathrm{ABC}$において，$(b+c):(c+a):(a+b)=4:5:6,R=\sqrt{3}$が成り立つとき，$\cos A,a,b,c$を求めよ．

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% 節末問題I4.2.3の解答（One More）★★★
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$(b+c):(c+a):(a+b)=4:5:6$であるから，$b+c=8k,c+a=10k,a+b=12k(k>0)\cdots(\mathrm{i})$とおくと，これらの辺々を足し合わせると，$2(a+b+c)=30k$

したがって，$a+b+c=15k\cdots(\mathrm{ii})$

(i)，(ii)より，$a=7k,b=5k,c=3k\cdots(\mathrm{iii})$

余弦定理より，$\cos A=\frac{(5k)^2+(3k)^2-(7k)^2}{2\cdot 5k\cdot 3k}=-\frac{1}{2}$

$\cos A=-\frac{1}{2}$であるから，$A=120^\circ$

これと$R=\sqrt{3}$であるから，正弦定理より，$\frac{a}{\sin 120^\circ}=2\sqrt{3}$

よって，$a=2\sqrt{3}\sin 120^\circ=3$

このとき，(iii)より，$k=\frac{3}{7}$であり，$b=\frac{15}{7},c=\frac{9}{7}$

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% 節末問題I4.2.4：（One More）★★★
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$\triangle\mathrm{ABC}$において，$\mathrm{AB}=x-1,\mathrm{AC}=x,\mathrm{BC}=x+1$のとき，次の問いに答えよ．

(1) $x$のとり得る値の範囲を求めよ．

(2) $\triangle\mathrm{ABC}$が鈍角三角形となる$x$の値の範囲を求めよ．

(3) $\triangle\mathrm{ABC}$の1つの内角が$120^\circ$であるとき，$x$の値，外接円の半径を求めよ．

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% 節末問題I4.2.4の解答（One More）★★★
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(1) $x-1<x<x+1$であるから，三角形が成立する条件より，
\[
x+1<x+(x-1)
\]

よって，$x$のとり得る値の範囲は，$x>2\cdots(\mathrm{i})$

(2) 辺$\mathrm{BC}$が最大の辺であるから，鈍角三角形となる条件は$A>90^{\circ}$，すなわち，$\cos A<0$である．

余弦定理より，

\[
\cos A=\frac{x^2+(x-1)^2-(x+1)^2}{2x(x-1)}<0
\]

したがって，$x^2+(x-1)^2-(x+1)^2<0$

整理すると，$x^2-4x<0$

ゆえに，$x(x-4)<0$

これを解くと，$0<x<4\cdots(\mathrm{ii})$

よって，(i)，(ii)より，求める$x$の値の範囲は，$2<x<4\cdots(\mathrm{iii})$

(3) 長さ$x+1$の辺に対する角が$120^{\circ}$になるから，余弦定理より，

\[
(x+1)^2=x^2+(x-1)^2-2\cdot x\cdot(x-1)\cdot\cos 120^{\circ}
\]

したがって，$2x^2-5x=0$であるから，$x(2x-5)=0$

(iii) より，$2<x<4$であるから$x=\frac{5}{2}$

外接円の半径を$R$とすると，正弦定理より，$\frac{x+1}{\sin 120^{\circ}}=2R$

よって，$R=\frac{1}{2}\cdot\frac{\frac{5}{2}+1}{\sin 120^\circ}=\frac{1}{2}\cdot\frac{7}{2}\div\frac{\sqrt{3}}{2}=\frac{7}{2\sqrt{3}}=\frac{7\sqrt{3}}{6}$

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% 節末問題I4.2.5：（One More）★★★
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次の等式が成り立つとき，$\triangle\mathrm{ABC}$はどのような三角形か．

\[
(a-b)\sin^2C=a\sin^2A-b\sin^2B
\]

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% 節末問題I4.2.5の解答（One More）★★★
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正弦定理より，

\[
(a-b)(\frac{c}{2R})^2=a(\frac{a}{2R})^2-b(\frac{b}{2R})^2
\]

両辺に$4R^2$を掛けると，

\[
(a-b)c^2=a^3-b^3
\]

したがって，$(a-b)c^2=(a-b)(a^2+ab+b^2)$

ゆえに，$(a-b)\{c^2-(a^2+ab+b^2)\}=0$

したがって，$a=b$または$c^2-(a^2+ab+b^2)=0$

(i) $a=b$のとき

$\triangle\mathrm{ABC}$は$\mathrm{BC}=\mathrm{CA}$の二等辺三角形である．

(ii) $c^2=a^2+ab+b^2$のとき

余弦定理より，

\[
\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-(a^2+ab+b^2)}{2ab}=\frac{-ab}{2ab}=-\frac{1}{2}
\]

したがって，$C=120^{\circ}$

よって，(i)，(ii)より，$\triangle\mathrm{ABC}$は$\mathrm{BC}=\mathrm{CA}$の二等辺三角形または$C=120^{\circ}$の三角形

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% 節末問題I4.3.1：（One More）★★
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半径$a$の円に内接する正$n$角形の面積，および外接する正$n$角形の面積を，それぞれ$a$と$n$を用いて表せ．

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% 節末問題I4.3.1の解答（One More）★★
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右の図のように，円Oに内接する正$n$角形の1辺をAB，外接する正$n$角形の1辺をCDとし，CDの中点をMとする．

$\mathrm{OA}=a,\mathrm{OB}=a,\angle\mathrm{AOB}=\frac{360^{\circ}}{n}$より，$\triangle\mathrm{OAB}$の面積は，

\[
\triangle\mathrm{OAB}=\frac{1}{2}a^2\sin\frac{360^{\circ}}{n}
\]

よって，内接する正$n$角形の面積は，

\[
n\times\triangle\mathrm{OAB}=\frac{1}{2}na^2\sin\frac{360^{\circ}}{n}
\]

また，$\mathrm{OM}=a,\mathrm{DM}=a\tan\frac{180^{\circ}}{n}$より，$\triangle\mathrm{OCD}$の面積は，

\[
\triangle\mathrm{OCD}=2\cdot\frac{1}{2}a\cdot a\tan\frac{180^{\circ}}{n}=a^2\tan\frac{180^{\circ}}{n}
\]

よって，外接する正$n$角形の面積は，

\[
n\times\triangle\mathrm{OCD}=na^2\tan\frac{180^{\circ}}{n}
\]

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% 節末問題I4.3.2：（One More）★★★
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$\triangle\mathrm{ABC}$において，$\frac{\sin A}{13}=\frac{\sin B}{8}=\frac{\sin C}{7}$が成り立つとする．このとき，次の問いに答えよ．

(1) $\cos A,\sin A$の値を求めよ．

(2) $\triangle\mathrm{ABC}$の内接円の半径が1のとき，$\triangle\mathrm{ABC}$の面積，$\triangle\mathrm{ABC}$の外接円の半径を求めよ．

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% 節末問題I4.3.2の解答（One More）★★★
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(1) 正弦定理$\frac{\mathrm{BC}}{\sin A}=\frac{\mathrm{CA}}{\sin B}=\frac{\mathrm{AB}}{\sin C}$より，

\[
\mathrm{BC}:\mathrm{CA}:\mathrm{AB}=\sin A:\sin B:\sin C\cdots(\mathrm{i})
\]

$\frac{\sin A}{13}=\frac{\sin B}{8}=\frac{\sin C}{7}$より，

\[
\sin A:\sin B:\sin C=13:8:7\cdots(\mathrm{ii})
\]

(i)，(ii)より，$\mathrm{BC}:\mathrm{CA}:\mathrm{AB}=13:8:7$

$\mathrm{BC}=13k,\mathrm{CA}=8k,\mathrm{AB}=7k(k>0)$とおくと，余弦定理より，

\[
\cos A=\frac{(8k)^2+(7k)^2-(13k)^2}{2\cdot 8k\cdot 7k}=-\frac{1}{2}
\]

$0^{\circ}<A<180^{\circ}$であるから，$\sin A>0$より，

\[
\sin A=\sqrt{1-\cos^2A}=\sqrt{1-(-\frac{1}{2})^2}=\frac{\sqrt{3}}{2}
\]

(2) $\triangle\mathrm{ABC}$の面積を$S$とする．

\[
S=\frac{1}{2}\cdot 8k\cdot 7k\cdot\sin A=14\sqrt{3}k^2
\]

また，内接円の半径が1であるから，

\[
S=\frac{1}{2}\cdot(13k+8k+7k)\cdot 1=14k
\]

したがって，$7k(\sqrt{3}k-1)=0$

$k>0$より，$k=\frac{1}{\sqrt{3}}$

ゆえに，$\mathrm{AB}=7k=7\cdot\frac{1}{\sqrt{3}}=\frac{7\sqrt{3}}{3}$

求める$\triangle\mathrm{ABC}$の面積$S$は，$S=14k=14\cdot\frac{1}{\sqrt{3}}=\frac{14\sqrt{3}}{3}$

外接円の半径を$R$とすると，正弦定理より，$\frac{13k}{\sin A}=2R$

よって，$\triangle\mathrm{ABC}$の外接円の半径は，

\[
R=\frac{13k}{2\sin A}=13\cdot\frac{1}{\sqrt{3}}\div(2\cdot\frac{\sqrt{3}}{2})=\frac{13}{3}
\]

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% 節末問題I4.3.3：（One More）★★★
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円に内接する四角形ABCDがある．$\mathrm{AB}=3,\mathrm{BC}=\mathrm{CD}=\sqrt{3}$,$\mathrm{DA}=2$とする．このとき，次の問いに答えよ．

(1) $\cos\angle\mathrm{BAD}$と対角線BDの長さを求めよ．

(2) 2つの対角線ACとBDの交点をEとする．$\mathrm{BE}:\mathrm{ED}$とBEの長さを求めよ．

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% 節末問題I4.3.3の解答（One More）★★★
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(1) 四角形ABCDは円に内接するから，

\[
\angle{\mathrm{BCD}}=180^\circ-\angle\mathrm{BAD}
\]

$\triangle\mathrm{ABD}$において，余弦定理より，

\[
\mathrm{BD}^2=3^2+2^2-2\cdot 3\cdot 2\cdot\cos\angle\mathrm{BAD}=13-12\cos\angle\mathrm{BAD}\cdots(\mathrm{i})
\]

$\triangle\mathrm{BCD}$において，余弦定理より，

\[
\mathrm{BD}^2=(\sqrt{3})^2+(\sqrt{3})^2-2\cdot\sqrt{3}\cdot\sqrt{3}\cdot\cos(180^{\circ}-\angle\mathrm{BAD})=6+6\cos\angle\mathrm{BAD}\cdots(\mathrm{ii})
\]

(i)，(ii)より，$13-12\cos\angle\mathrm{BAD}=6+6\cos\angle\mathrm{BAD}$

したがって，$\cos\angle\mathrm{BAD}=\frac{7}{18}$

(i)に代入すると，$\mathrm{BD}^2=13-12\cdot(\frac{7}{18})=\frac{25}{3}$

よって，$\mathrm{BD}>0$より，$\mathrm{BD}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}$

(2) $\mathrm{BE}:\mathrm{ED}=\triangle\mathrm{ABC}:\triangle\mathrm{ACD}=\mathrm{AB}\cdot\mathrm{BC}:\mathrm{CD}\cdot\mathrm{DA}=3\cdot\sqrt{3}:\sqrt{3}\cdot 2=3:2$

よって，$\mathrm{BE}=\mathrm{BD}\cdot\frac{3}{3+2}=\frac{5\sqrt{3}}{3}\cdot\frac{3}{5}=\sqrt{3}$

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% 節末問題I4.3.4：（One More）★★★
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$\mathrm{AB}=\sqrt{6},\mathrm{AD}=\sqrt{3},\mathrm{AE}=1$である右の図のような直方体$\mathrm{ABCD}-\mathrm{EFGH}$がある．このとき，次の値を求めよ．

(1) $\angle\mathrm{ACF}$

(2) $\triangle\mathrm{ACF}$の面積

(3) 四面体$\mathrm{BAFC}$の体積

(4) Bから平面$\mathrm{AFC}$に下ろした垂線の長さ

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% 節末問題I4.3.4の解答（One More）★★★
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(1) $\mathrm{AC}^2=\mathrm{AB}^2+\mathrm{BC}^2$より，$\mathrm{AC}=\sqrt{6+3}=\sqrt{9}=3$

$\mathrm{AF}^2=\mathrm{AE}^2+\mathrm{EF}^2$より，$\mathrm{AF}=\sqrt{6+1}=\sqrt{7}$

$\mathrm{CF}^2=\mathrm{BF}^2+\mathrm{BC}^2$より，$\mathrm{CF}=\sqrt{1+3}=\sqrt{4}=2$

$\triangle\mathrm{AFC}$において，余弦定理より，

\[
\cos\angle\mathrm{ACF}=\frac{3^2+2^2-(\sqrt{7})^2}{2\cdot 3\cdot 2}=\frac{1}{2}
\]

よって，$0^{\circ}<\angle\mathrm{ACF}<180^{\circ}$より，$\angle\mathrm{ACF}=60^\circ$

(2) $\triangle\mathrm{ACF}$の面積を$S$とすると，$S=\frac{1}{2}\cdot 3\cdot 2\cdot\sin\angle\mathrm{ACF}=\frac{1}{2}\cdot 3\cdot 2\cdot\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$

(3) 四面体$\mathrm{BAFC}$の体積を$V$とすると，

\[
V=\frac{1}{3}\cdot\triangle\mathrm{ABF}\cdot\mathrm{BC}=\frac{1}{3}\cdot(\frac{1}{2}\cdot\sqrt{6}\cdot 1)\cdot\sqrt{3}=\frac{\sqrt{18}}{6}=\frac{\sqrt{2}}{2}\cdots(\mathrm{i})
\]

(4) 求める垂線の長さを$h$とすると，$V=\frac{1}{3}\cdot\triangle\mathrm{ACF}\cdot h\cdots(\mathrm{ii})$

したがって，(i)，(ii)より，$\frac{\sqrt{2}}{2}=\frac{1}{3}\cdot\frac{3\sqrt{3}}{2}\cdot h$

よって，$h=\frac{\sqrt{6}}{3}$

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% 節末問題I4.3.5：（One More）★★★
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底面の半径$\sqrt{5}$の直円錐に半径1の球が内接している．このとき，この直円錐の体積を求めよ．

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% 節末問題I4.3.5の解答（One More）★★★
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右の図のように，直円錐の頂点Aと底面の円の直径BCを含む平面（切断面）を考える．球の中心をOとし，切断面において，内接円が辺AB，AC，BCと接する点を，それぞれ$\mathrm{D},\mathrm{E},\mathrm{F}$とし，$\mathrm{AD}=x,\mathrm{AO}=y$とおく．

$\triangle\mathrm{ADO}\sim\triangle\mathrm{AFB}$より，

$\mathrm{AD}:\mathrm{AF}=\mathrm{DO}:\mathrm{FB}$，すなわち，$x:(y+1)=1:\sqrt{5}\cdots(\mathrm{i})$

$\mathrm{AO}:\mathrm{AB}=\mathrm{DO}:\mathrm{FB}$，すなわち，$y:(x+\sqrt{5})=1:\sqrt{5}\cdots(\mathrm{ii})$

(i)より，$y+1=\sqrt{5}x\cdots(\mathrm{iii})$

(ii)より，$x+\sqrt{5}=\sqrt{5}y$，すなわち，$x=\sqrt{5}(y-1)\cdots(\mathrm{iv})$

(iii)，(iv)より，$y+1=\sqrt{5}\cdot\sqrt{5}(y-1)$

したがって，$y+1=5(y-1)$より，$y=\frac{3}{2}$

このとき，これを(iv)に代入すると，$x=\frac{\sqrt{5}}{2}$

よって，求める直円錐の体積を$V$とすると，$V=\frac{1}{3}\cdot\pi\cdot(\sqrt{5})^2\cdot(\frac{3}{2}+1)=\frac{25}{6}\pi$

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% 章末問題I4.1：（One More）★★
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$\sin\theta+\cos\theta=\frac{1}{\sqrt{5}}$のとき，$\tan^3\theta+\frac{1}{\tan^3\theta}$の値を求めよ．

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% 章末問題I4.1の解答（One More）★★
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\begin{align*}
\tan^3\theta+\frac{1}{\tan^3\theta}=&(\tan\theta+\frac{1}{\tan\theta})^3-3\tan\theta\cdot\frac{1}{\tan\theta}(\tan\theta+\frac{1}{\tan\theta})\\
=&(\tan\theta+\frac{1}{\tan\theta})^3-3(\tan\theta+\frac{1}{\tan\theta})\cdots(\mathrm{i})
\end{align*}

また，

\[
\tan\theta+\frac{1}{\tan\theta}=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}\cdots(\mathrm{ii})
\]

$\sin\theta+\cos\theta=\frac{1}{\sqrt{5}}$の両辺を2乗すると，$\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{5}$

したがって，$1+2\sin\theta\cos\theta=\frac{1}{5}$

整理すると，$\sin\theta\cos\theta=-\frac{2}{5}$

ゆえに，(ii)より，$\tan\theta+\frac{1}{\tan\theta}=-\frac{5}{2}$

よって，(i)より，$\tan^3\theta+\frac{1}{\tan^3\theta}=(-\frac{5}{2})^3-3\cdot(-\frac{5}{2})=-\frac{65}{8}$

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% 章末問題I4.2：（One More）★★★★
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$\sin^2\theta+2a\cos\theta-3=0$が$90^{\circ}\leqq\theta\leqq 180^{\circ}$の範囲に解をもつための定数$a$の値の範囲を求めよ．

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% 章末問題I4.2の解答（One More）★★★★
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$\cos\theta=t$とおくと，与えられた方程式は，$1-t^2+2at-3=0$，すなわち，$t^2-2at+2=0$となり，この2次方程式の判別式を$D$とする．

$90^{\circ}\leqq\theta\leqq 180^{\circ}$のとき，$-1\leqq\cos\theta\leqq 0$より，$-1\leqq t\leqq 0$

$f(t)=t^2-2at+2$とおくと，与えられた方程式が$90^{\circ}\leqq\theta\leqq 180^{\circ}$に解をもつのは，$y=f(t)$のグラフが$-1\leqq t\leqq 0$の範囲で$t$軸と共有点をもつときである．

また，$f(-1)=2a+3,f(0)=2$より，$f(0)>0$であるから，求める条件は，(i)，(ii)の場合に分けられる．

(i) $f(-1)\leqq 0$，すなわち，$a\leqq-\frac{3}{2}$のとき

$f(-1)\cdot f(0)\leqq 0$となることから，これは求める条件を満たす．

(ii) $f(-1)>0$，すなわち，$a>-\frac{3}{2}$のとき

求める条件は，（ア）$D\geqq 0$（イ）軸$t=a$が$-1<t<0$である．

（ア）$\frac{D}{4}=(-a)^2-1\cdot 1\cdot 2=a^2-2$であり，$D\geqq 0$から，$a\leqq-\sqrt{2},\sqrt{2}\leqq a$

（イ）$y=f(t)$の軸は直線$t=a$であり，軸が$-1<t<0$の間にあるから，$-1<a<0$

したがって，（ア），（イ）より，これらを同時に満たす$a$の値は存在しない．

よって，(i)，(ii)より，求める定数$a$の値の範囲は，$a\leqq-\frac{3}{2}$

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% 章末問題I4.3：（One More）★★★
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三角形ABCの辺BCを$8:5$に内分する点をDとする．$\mathrm{AB}=7,\mathrm{AC}=5\sqrt{3},\mathrm{AD}=5$であるとき，次の値を求めよ．

(1) $\angle\mathrm{ADB}$の大きさ

(2) $\triangle\mathrm{ABC}$の面積

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% 章末問題I4.3の解答（One More）★★★
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(1) $\mathrm{BD}:\mathrm{DC}=8:5$より，$\mathrm{BD}=8k,\mathrm{DC}=5k(k>0)$，また，$\angle\mathrm{ADB}=\theta$とおく．

$\triangle\mathrm{ABD}$において，余弦定理より，$7^2=5^2+(8k)^2-2\cdot 5\cdot 8k\cdot\cos\theta$

整理すると，$8k^2-10k\cos\theta-3=0\cdots(\mathrm{i})$

$\triangle\mathrm{ACD}$において，余弦定理より，$({5\sqrt{3}})^2=5^2+(5k)^2-2\cdot 5\cdot 5k\cdot\cos(180^{\circ}-\theta)$

したがって，$75=25+25k^2+50k\cos\theta$

整理すると，$5k^2+10k\cos\theta-10=0\cdots(\mathrm{ii})$

(i)と(ii)の辺々を足し合わせると，$13k^2-13=0$

ゆえに，$k>0$より，$k=1$

これを(i)に代入すると，$\cos\theta=\frac{1}{2}$より，$\theta=60^\circ$

よって，$\angle\mathrm{ADB}=60^\circ$

(2) 求める$\triangle\mathrm{ABC}$の面積は，

\[
\triangle\mathrm{ABC}=\frac{13}{8}\triangle\mathrm{ABD}=\frac{13}{8}\cdot\frac{1}{2}\cdot 5\cdot 8\sin 60^{\circ}=\frac{65\sqrt{3}}{4}
\]

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% 章末問題I4.4：（One More）★★★★
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1辺の長さが3の正三角形ABCがある．辺AB，AC上に，それぞれ頂点とは異なる点D，Eを，$\mathrm{AD}=\mathrm{CE}$を満たすようにとる．また，四角形DBCEの面積を$S$とする．

(1) DEの長さの最小値を求めよ．

(2) 面積$S$の最小値とそのときのADの長さを求めよ．

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% 章末問題I4.4の解答（One More）★★★★
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(1) $\mathrm{AD}=\mathrm{CE}=x$とすると，

\[
0<x<3
\]

$\triangle\mathrm{ADE}$において，余弦定理より，

\begin{align*}
\mathrm{DE}^2=&x^2+(3-x)^2-2x\cdot(3-x)\cdot\cos 60^{\circ}\\
=&3x^2-9x+9\\
=&3(x-\frac{3}{2})^2+\frac{9}{4}
\end{align*}

したがって，$0<x<3$の範囲において，$\mathrm{DE}^2$は$x=\frac{3}{2}$のとき，最小値$\frac{9}{4}$

よって，$\mathrm{DE}>0$であるから，DEは$\mathrm{AD}=\frac{3}{2}$のとき，最小値$\frac{3}{2}$

(2)

\begin{align*}
S&=\triangle\mathrm{ABC}-\triangle\mathrm{ADE}\\
&=\frac{1}{2}\cdot 3\cdot\frac{3\sqrt{3}}{2}-\frac{1}{2}x(3-x)\sin 60^{\circ}\\
&=\frac{\sqrt{3}}{4}\{9-x(3-x)\}\\
&=\frac{\sqrt{3}}{4}(x^2-3x+9)\\
&=\frac{\sqrt{3}}{4}(x-\frac{3}{2})^2+\frac{27\sqrt{3}}{16}
\end{align*}

よって，$0<x<3$の範囲において，$S$は$\mathrm{AD}=\frac{3}{2}$のとき，最小値$\frac{27\sqrt{3}}{16}$

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% 章末問題I4.5：（One More）★★★
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1辺の長さが6の正四面体ABCDについて，辺BC上に$\mathrm{BE}:\mathrm{EC}=1:2$となるように点Eをとり，辺CDの中点をM，$\angle\mathrm{EAM}=\theta$とする．このとき，次の値を求めよ．

(1) $\cos\theta$

(2) $\triangle\mathrm{AEM}$の面積$S$

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% 章末問題I4.5の解答（One More）★★★
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(1) $\triangle\mathrm{ACM}$は，$\angle\mathrm{AMC}=90^{\circ}$の直角三角形であるから，
\[
\mathrm{AM}=\mathrm{AC}\sin\angle\mathrm{ACM}=\mathrm{AC}\sin 60^{\circ}=6\cdot\frac{\sqrt{3}}{2}={3\sqrt{3}}
\]

$\triangle\mathrm{ABE}$において，余弦定理により，

\[
\mathrm{AE}^2=6^2+2^2-2\cdot 6\cdot 2\cdot\cos 60^{\circ}=28
\]

したがって，$\mathrm{AE}>0$より，$\mathrm{AE}=2\sqrt{7}$

$\triangle\mathrm{CEM}$において，余弦定理より，

\begin{align*}
\mathrm{EM}^2&=\mathrm{CE}^2+\mathrm{CM}^2-2\cdot\mathrm{CE}\cdot\mathrm{CM}\cdot\cos\angle\mathrm{ECM}\\
&=4^2+3^2-2\cdot 4\cdot 3\cdot\cos 60^{\circ}=13
\end{align*}

ゆえに，$\mathrm{EM}>0$より，$\mathrm{EM}=\sqrt{13}$

$\triangle{\mathrm{EAM}}$において，余弦定理より，

\[
\cos\theta=\frac{(3\sqrt{3})^2+(2\sqrt{7})^2-(\sqrt{13})^2}{2\cdot 3\sqrt{3}\cdot 2\sqrt{7}}=\frac{\sqrt{21}}{6}
\]

(2) $\sin\theta>0$であるから，

\[
\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{\sqrt{21}}{6})^2}=\frac{\sqrt{15}}{6}
\]

よって，求める$\triangle\mathrm{AEM}$の面積$S$は，

\[
S=\frac{1}{2}\cdot\mathrm{AM}\cdot\mathrm{AE}\cdot\sin\theta=\frac{1}{2}\cdot 3\sqrt{3}\cdot 2\sqrt{7}\cdot\frac{\sqrt{15}}{6}=\frac{3\sqrt{35}}{2}
\]

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% 節末問題I4.1.1の側注：（One More）★★★
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% 側注1
$\mathrm{AB}^2=\mathrm{AC}^2+\mathrm{BC}^2$

% 側注2
\begin{align*}
&\mathrm{BC}:\mathrm{CD}:\mathrm{BD}\\
=&1:1:\sqrt{2}
\end{align*}

% 側注3
この2重根号は外すことができない．

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% 節末問題I4.1.2の側注：（One More）★★★
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% 側注1
通分する．

% 側注2
$\sin^2\theta+\cos^2\theta=1$

% 側注3
$\theta$は鋭角，$\tan\theta=\frac{5}{12}$であることから，直角三角形をかき，三平方の定理を用いて解くこともできる．

% 側注4
$\tan\theta=\frac{\sin\theta}{\cos\theta}$より，$\sin\theta=\tan\theta\cos\theta$

% 側注5
$\theta$が鋭角$(0^{\circ}<\theta<90^{\circ})$のとき，$\cos\theta>0$であることに注意すること．

% 側注6
$1+\cos\theta=1+\frac{12}{13}=\frac{25}{13},1-\cos\theta=1-\frac{12}{13}=\frac{1}{13}$

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% 節末問題I4.1.3の側注：（One More）★★
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% 側注1
$\theta=90^\circ$のとき，$\sin\theta=1$

% 側注2
$\cos\theta\neq 0$より，両辺を$\cos\theta$で割ることができる．

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% 節末問題I4.1.4の側注：（One More）★★★
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% 側注1
$\tan\theta=\frac{\sin\theta}{\cos\theta},\sin^2\theta+\cos^2\theta=1$

% 側注2
$\sin^2\theta+\cos^2\theta=1$

% 側注3
$\sin^2\theta+\cos^2\theta=1$

% 側注4
$0^{\circ}<\theta<45^{\circ}$のとき，下の図のような直角三角形において，$a<b$

よって，$c\sin\theta<c\cos\theta$

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% 節末問題I4.2.1の側注：（One More）★★
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% 側注1
$c^2-a^2=b^2-ab,c^2-b^2=a^2-ab$を代入する．

% 側注2
約分する．

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% 節末問題I4.2.2の側注：（One More）★★★
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% 側注1
$C$の値が2つ求められたから，場合分けをして考える．

% 側注2
$\sin 75^\circ$は扱いにくいことから，正弦定理ではなく，余弦定理を用いる．

% 側注3
$(\sqrt{3}+1)^2-(\sqrt{6})^2=2(\sqrt{3}-1)>0$より，$\sqrt{3}+1>\sqrt{6}$

% 側注4
$2^2-(\sqrt{3}-1)^2=2\sqrt{3}>0$より，$\sqrt{3}-1<2$

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% 節末問題I4.2.3の側注：（One More）★★★
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% 側注1
$b+c=4k,c+a=5k,a+b=6k$とおいてもよいが，$a+b+c=\frac{15}{2}k$となり計算に手間が掛かる．そこで，計算を簡単にするために，$4:5:6$を$8:10:12$と考えている．

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% 節末問題I4.2.4の側注：（One More）★★★
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% 側注1
最大の辺の長さは，他の2辺の長さの和より小さい．

% 側注2
鈍角になることができるのは，最大の角のみである．

% 側注3
(i)より，分母が正であるから，分子は負である．

% 側注4
1つの内角が$120^\circ$であるから，$\triangle\mathrm{ABC}$は鈍角三角形である．

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% 節末問題I4.2.5の側注：（One More）★★★
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% 側注1
$\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}$

% 側注2
\begin{align*}
&a^3-b^3\\
=&(a-b)(a^2+ab+b^2)
\end{align*}

% 側注3
$\alpha\beta=0$ならば$\alpha=0$または$\beta=0$となる．

% 側注4
$ab\neq 0$

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% 節末問題I4.3.1の側注：（One More）★★
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% 側注1
正$n$角形を$n$個の合同な三角形に分割する．

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% 節末問題I4.3.2の側注：（One More）★★★
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% 側注1
$\sin^2A+\cos^2A=1$

% 側注2
$2R=\frac{\mathrm{BC}}{\sin A}$より，

\[
R=\frac{\mathrm{BC}}{2\sin A}
\]

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% 節末問題I4.3.3の側注：（One More）★★★
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% 側注1
$\angle\mathrm{BCD}+\angle\mathrm{BAD}=180^\circ$

% 側注2
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注3
下の線分比と面積比の関係を用いる．

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% 節末問題I4.3.4の側注：（One More）★★★
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% 側注1
三平方の定理を用いる．

% 側注2
$\triangle\mathrm{ABF}$を底面とすると，$V$が求められる．

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% 節末問題I4.3.5の側注：（One More）★★★
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% 側注1
$\sim$は相似を表す．$\angle\mathrm{DAO}=\angle\mathrm{FAB}$，$\angle\mathrm{ADO}=\angle\mathrm{AFB}=90^\circ$より，相似である．

% 側注2
2つの接線の長さは等しいので，$\mathrm{BD}=\mathrm{BF}=\sqrt{5}$である．（数学Aで学習する）

% 側注3
高さ$\mathrm{AF}=\mathrm{AO}+\mathrm{OF}$

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% 章末問題I4.1の側注：（One More）★★
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% 側注1
\begin{align*}
&a^3+b^3\\
=&(a+b)(a^2-ab+b^2)
\end{align*}

% 側注2
$\sin^2\theta+\cos^2\theta=1$

% 側注3
$\sin^2\theta+\cos^2\theta=1$

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% 章末問題I4.2の側注：（One More）★★★★
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% 側注1
$\sin^2\theta+\cos^2\theta=1$より，$\sin^2\theta=1-\cos^2\theta$

% 側注2
下の図のようになる$a$の値は存在しない．

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% 章末問題I4.3の側注：（One More）★★★
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% 側注1
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注2
(ii)に代入してもよい．

% 側注3
$\mathrm{BD}:\mathrm{BC}=8:13$より，$\triangle\mathrm{ABD}:\triangle\mathrm{ABC}=8:13$

なお，AからBCに垂線を下ろし，$60^\circ$の三角比から高さを考えることで面積を求めてもよい．

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% 章末問題I4.4の側注：（One More）★★★★
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% 側注1
1辺の長さが3であるから，$x<3$であり，また，辺の長さは正であるから，$0<x$である．

% 側注2
$\triangle\mathrm{ABC}$の面積は，$\frac{1}{2}\cdot 3\cdot\frac{3\sqrt{3}}{2}$

% 側注3
$\mathrm{DE}^2$が最小となるとき，$\mathrm{DE}$も最小となる．

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% 章末問題I4.5の側注：（One More）★★★
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% 側注1
$0^{\circ}<\theta<180^{\circ}$のとき，$\sin\theta>0$