【数学I】例題4.1.2：三角比を用いた測量（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.2：三角比を用いた測量（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

木に向かって水平な道路をまっすぐ歩いている人が，木の最上部にある点Pを見上げたところ，A地点でその仰角が$45^{\circ}$であった．その後，A地点から10 m進んでB地点に到達したとき，再び仰角を測ると$60^{\circ}$であった．この人の目の高さが地面から1.5 mであるとき，この木の高さを求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.2の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

右の図のように，$\mathrm{A}^{\prime},\mathrm{B}^{\prime},\mathrm{C},\mathrm{C}^\prime,\mathrm{P}$とおき，$\mathrm{BC}=\mathrm{B}^{\prime}\mathrm{C}^{\prime}=x\mathrm{~m},\mathrm{C}^{\prime}\mathrm{P}=y\mathrm{~m}$とする．

直角三角形$\mathrm{A}^{\prime}\mathrm{C}^{\prime}\mathrm{P}$において，

\[
y=(10+x)\tan{45}^\circ
\]

したがって，$y=10+x\cdots(\mathrm{i})$

直角三角形$\mathrm{B}^{\prime}\mathrm{C}^{\prime}\mathrm{P}$において，

\[
y=x\tan 60^{\circ}
\]

したがって，$y=\sqrt{3}x\cdots(\mathrm{ii})$

(i)，(ii)より，$\sqrt{3}x=10+x$

$(\sqrt{3}-1)x=10$より，$x=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

これと(i)より，$y=15+5\sqrt{3}$

ゆえに，$\mathrm{PC}=y+1.5=16.5+5\sqrt{3}$

よって，木の高さは，$16.5+5\sqrt{3}(\mathrm{m})$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.2：三角比を用いた測量（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

木に向かって水平な道路をまっすぐ歩いている人が，木の最上部にある点Pを見上げたところ，A地点でその仰角が$30^{\circ}$であった．その後，A地点から10 m進んでB地点に到達したとき，再び仰角を測ると$45^{\circ}$であった．この人の目の高さが地面から1.5 mであるとき，この木の高さを求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.2の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

右の図のように，$\mathrm{A}^{\prime},\mathrm{B}^{\prime},\mathrm{C},\mathrm{C}^\prime,\mathrm{P}$とおき，$\mathrm{BC}=\mathrm{B}^{\prime}\mathrm{C}^{\prime}=x\mathrm{~m},\mathrm{C}^{\prime}\mathrm{P}=y\mathrm{~m}$とする．

直角三角形$\mathrm{A}^{\prime}\mathrm{C}^{\prime}\mathrm{P}$において，

\[
y=(10+x)\tan{30}^\circ
\]

したがって，$y=\frac{10+x}{\sqrt{3}}\cdots(\mathrm{i})$

直角三角形$\mathrm{B}^{\prime}\mathrm{C}^{\prime}\mathrm{P}$において，

\[
y=x\tan 45^{\circ}
\]

したがって，$y=x\cdots(\mathrm{ii})$

(i)，(ii)より，$\sqrt{3}x=10+x$

$(\sqrt{3}-1)x=10$より，$x=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

これと(ii)より，$y=5+5\sqrt{3}$

ゆえに，$\mathrm{PC}=y+1.5=6.5+5\sqrt{3}$

よって，木の高さは，$6.5+5\sqrt{3}(\mathrm{m})$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.2の側注：三角比を用いた測量（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\tan{45}^\circ=\frac{y}{10+x}$

% 側注2
$\tan{45}^\circ=1$

% 側注3
$\tan{60}^\circ=\frac{y}{x}$

% 側注4
$\tan{60}^\circ=\sqrt{3}$

% 側注5
\begin{align*}
&\frac{10}{\sqrt{3}-1}\\
=&\frac{10(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\
=&\frac{10(\sqrt{3}+1)}{2}
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.2の側注：三角比を用いた測量（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\tan{30}^\circ=\frac{y}{10+x}$

% 側注2
$\tan{30}^\circ=\frac{1}{\sqrt{3}}$

% 側注3
$\tan{45}^\circ=\frac{y}{x}$

% 側注4
$\tan{45}^\circ=1$

% 側注5
\begin{align*}
&\frac{10}{\sqrt{3}-1}\\
=&\frac{10(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\
=&5(\sqrt{3}+1)
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.2の考え方・補足など：三角比を用いた測量（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 考え方
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

与えられた値をもとに直角三角形をかき，三平方の定理や三角比の利用を考える．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 注意
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

右の図のように，点Aから点Pを見るとき，点Aを通る水平線とAPとのなす角を仰（ぎょう）角という．また，点Aから点$\mathrm{P}'$を見るとき，点Aを通る水平線と$\mathrm{AP}'$とのなす角を俯（ふ）角という．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 注意
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

例題において$\sqrt{3}=1.732$とし，木の高さの小数第2位を四捨五入すると，$16.5+5\sqrt{3}\fallingdotseq 25.2$となる（近似の記号$\fallingdotseq$は$\approx$と表すこともある）．例題では指示がなかったが，問題によっては「四捨五入して求めよ」などと指示がある場合があるので注意すること．また，右の図のような$30^{\circ},45^{\circ},60^{\circ}$の三角比は覚えて，自在に利用できるようになるとよい．