問題の解答
検索用コード（LaTeX）
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.3：$15^\circ$の三角比（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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) $\sin 15^{\circ},\cos 15^{\circ},\tan 15^{\circ}$の値を求めよ．
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.3の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1) $\angle\mathrm{CBD}=180^\circ-(90^\circ+30^{\circ})=60^\circ$より，$\mathrm{CD=\sqrt{3},\mathrm{BD}=2}$
また，$\angle\mathrm{ABC}=180^{\circ}-(90^\circ+15^{\circ})=75^{\circ}$より，$\angle\mathrm{ABD}=75^{\circ}-60^{\circ}=15^{\circ}$
したがって，$\triangle\mathrm{ABD}$は$\mathrm{AD}=\mathrm{BD}$の二等辺三角形である．
ゆえに，$\mathrm{CD}=\sqrt{3}$より，$\mathrm{AC}=\mathrm{AD}+\mathrm{CD}=2+\sqrt{3}\cdots(\mathrm{i})$
$\triangle\mathrm{ABC}$において，三平方の定理より，
\[
\mathrm{AB}^2=(2+\sqrt{3})^2+1^2=8+4\sqrt{3}
\]
よって，$\mathrm{AB}>0$より，
\[
\mathrm{AB}=\sqrt{8+4\sqrt{3}}=\sqrt{8+2\sqrt{12}}=\sqrt{6}+\sqrt{2}\cdots(\mathrm{ii})
\]
(2) $\triangle\mathrm{ABC}$において，(i)，(ii)，$\mathrm{BC}=1$より，
\begin{align*}
&\sin 15^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{\sqrt{6}-\sqrt{2}}{4}\\
&\cos 15^{\circ}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{2+\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{(2+\sqrt{3})(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{\sqrt{6}+\sqrt{2}}{4}\\
&\tan 15^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}
\end{align*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.3：$15^\circ$の三角比（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
二等辺三角形ABCにおいて$\mathrm{AB}=\mathrm{AC},\mathrm{BC}=1,\angle\mathrm{A}=36^{\circ}$とし，$\angle\mathrm{B}$の二等分線と辺ACの交点をDとするとき，次の値を求めよ．
(1) 辺BDの長さ
(2) 辺ABの長さ
(3) $\sin 18^{\circ}$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.3の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1) $\triangle\mathrm{ABC}$は二等辺三角形であり，$\angle\mathrm{A}=36^{\circ}$であるから，
\[
\angle\mathrm{B}=\angle\mathrm{C}=\frac{180^{\circ}-36^{\circ}}{2}=72^{\circ}
\]
BDは$\angle\mathrm{B}$の二等分線であるから，
\[
\angle\mathrm{ABD}=\angle\mathrm{CBD}=36^{\circ}
\]
また，$\angle\mathrm{BCD}=\angle\mathrm{BDC}=72^{\circ}$であるから，$\triangle\mathrm{BCD}$は$\mathrm{BC}=\mathrm{BD}$の二等辺三角形である．
よって，$\mathrm{BD}=\mathrm{BC}=1$
(2) $\triangle\mathrm{DAB}$は$\mathrm{DA}=\mathrm{DB}$を満たす二等辺三角形であるから，$\mathrm{AD}=1$
$\mathrm{AB}=x$とすると，$\mathrm{CD}=x-1$
$\triangle\mathrm{ABC}\backsim\triangle\mathrm{BCD}$であるから，$\mathrm{AB}:\mathrm{BC}=\mathrm{BC}:\mathrm{CD}$，すなわち，$x:1=1:(x-1)$
したがって，$x(x-1)=1$
整理すると，$x^2-x-1=0$
これを解くと，$x>0$より，$x=\frac{1+\sqrt{5}}{2}$
よって，$\mathrm{AB}=\frac{1+\sqrt{5}}{2}$
(3)
\[
\sin 18^{\circ}=\frac{\frac{1}{2}\mathrm{BC}}{\mathrm{AB}}=\frac{1}{2x}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{(\sqrt{5}+1)(\sqrt{5}-1)}=\frac{\sqrt{5}-1}{4}
\]
あわせて読みたい
【数学I】4章：図形と計量（基本事項）
検索用コード（LaTeX） 本文・解答側注 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % 基本事項I4.1.1：三角比（One More） %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%...
あわせて読みたい
【数学I】4章：図形と計量（節末問題・章末問題）
節末I4.1.1〜I4.1.5の解答 節末I4.1.1節末I4.1.2節末I4.1.3節末I4.1.4節末I4.1.5 リンク（関連例題） https://onemath.net/onemorei-reidai4-1-3 https://onemath.net/o...