問題の解答
検索用コード（LaTeX）
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% 例題I4.3.8：空間図形の測量（One More）★★
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水平な地面に垂直に立つ木があり，木の頂点をA，その真下の地面上の点をDとする．また，地面上で互いに80 m離れた2点B，Cを定め，$\angle\mathrm{ABC},\angle\mathrm{ACB},\angle\mathrm{ACD}$を測定したところ，それぞれ$75^{\circ},60^{\circ},30^{\circ}$であった．このとき，木の高さADを求めよ．
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% 例題I4.3.8の解答（One More）★★
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$\triangle\mathrm{ABC}$において，
\[
\angle\mathrm{BAC}=180^{\circ}-(75^{\circ}+60^{\circ})=45^{\circ}
\]
正弦定理より，
\[
\frac{80}{\sin 45^{\circ}}=\frac{\mathrm{AB}}{\sin 60^{\circ}}
\]
したがって，
\[
\mathrm{AB}=\frac{80}{\sin 45^\circ}\cdot\sin 60^\circ=80\div\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}=40\sqrt{6}
\]
$\mathrm{AC}=x$とすると，余弦定理より，
\[
(40\sqrt{6})^2=80^2+x^2-2\cdot 80\cdot x\cdot\cos 60^{\circ}
\]
ゆえに，$x^2-80x-3200=0$
これを解くと，$x=40\pm 40\sqrt{3}$
したがって，$x>0$より，$x=40+40\sqrt{3}$
すなわち，$\mathrm{AC}=40+40\sqrt{3}$
$\triangle\mathrm{ACD}$において，
\[
\mathrm{AD}=\mathrm{AC}\cdot\sin 30^{\circ}=(40+40\sqrt{3})\cdot\frac{1}{2}=20+20\sqrt{3}
\]
よって，木の高さは，$\mathrm{AD}=20+20\sqrt{3}(\mathrm{m})$
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% 問題I4.3.8：空間図形の測量（One More）★★
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水平な地面に垂直に立つ木があり，木の頂点をA，その真下の地面上の点をDとする．また，地面上で互いに200 m離れた2点B，Cを定め，$\angle\mathrm{ABC},\angle\mathrm{ACB},\angle\mathrm{ACD}$を測定したところ，それぞれ$75^{\circ},60^{\circ},60^{\circ}$であった．このとき，木の高さADを求めよ．
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% 問題I4.3.8の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\triangle\mathrm{ABC}$において，
\[
\angle\mathrm{BAC}=180^{\circ}-(75^{\circ}+60^{\circ})=45^{\circ}
\]
正弦定理より，
\[
\frac{200}{\sin 45^{\circ}}=\frac{\mathrm{AB}}{\sin 60^{\circ}}
\]
したがって，
\[
\mathrm{AB}=\frac{200}{\sin 45^\circ}\cdot\sin 60^\circ=200\div\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}=100\sqrt{6}
\]
$\mathrm{AC}=x$とすると，余弦定理より，
\[
(100\sqrt{6})^2=200^2+x^2-2\cdot 200\cdot x\cdot\cos 60^{\circ}
\]
ゆえに，$x^2-200x-20000=0$
これを解くと，$x=100\pm 100\sqrt{3}$
したがって，$x>0$より，$x=100+100\sqrt{3}$
すなわち，$\mathrm{AC}=100+100\sqrt{3}$
$\triangle\mathrm{ACD}$おいて，
\[
\mathrm{AD}=\mathrm{AC}\cdot\sin 60^{\circ}=(100+100\sqrt{3})\cdot\frac{\sqrt{3}}{2}=50\sqrt{3}+150
\]
よって，木の高さは，$\mathrm{AD}=50\sqrt{3}+150(\mathrm{m})$
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