【高校数学Ⅰ】3-4-3 空間図形への応用｜要点まとめ

\(∠AHB=180°-(60°+75°)=45°\)
\(△ABH\)において、正弦定理より、
\(\displaystyle \frac{300}{\sin45^{\circ}}=\frac{HB}{\sin60^{\circ}}\)
\(\displaystyle HB=\frac{300}{\sin45^{\circ}}\times\sin60^{\circ}\)
\(\displaystyle HB=150\sqrt{6}\) \(\displaystyle \tan30^{\circ}=\frac{PH}{150\sqrt{6}}\)
\(\displaystyle PH=150\sqrt{6}\times\frac{1}{\sqrt{3}}\)
\(\displaystyle PH=150\sqrt{2}\)
よって、\(150\sqrt{2}\)m

\(△ABM\)において、三平方の定理より、
\(AM^2=10^2-5^2\)
\(AM^2=75\)
\(AM>0\)より、
\(AM=5\sqrt{3}\)
\(△AMB\)において、余弦定理より、
\(\cos∠ADM\)
\(\displaystyle =\frac{10^2+(5\sqrt{3})^2-(5\sqrt{3})^2}{2\times10\times5\sqrt{3}}\)
\(\displaystyle =\frac{100}{100\sqrt{3}}\)
\(\displaystyle =\frac{\sqrt{3}}{3}\)

\(\displaystyle \sin∠ADM=\sqrt{1-\left(\frac{\sqrt{3}}{3}\right)^2}=\frac{\sqrt{6}}{3}\)
よって、
\(\displaystyle S=\frac{1}{2}\times10\times5\sqrt{3}\sin∠ADM\)
\(\displaystyle \ \ =25\sqrt{3}\times\frac{\sqrt{6}}{3}\)
\(\ \ =25\sqrt{2}\)cm\(^2\)

\(△ADM\)において、
\(\displaystyle \sin∠ADM=\frac{h}{AD}\)
\(h=AD\sin∠ADM\)
\(\displaystyle \ \ =10\times\frac{\sqrt{6}}{3}\)
\(\displaystyle \ \ =\frac{10\sqrt{6}}{3}\)cm

\(△BCD\)の底面積\(S\)は、
\(\displaystyle S=\frac{1}{2}\times10\times10\sin60^{\circ}\)
\(\ \ =25\sqrt{3}\)cm\(^2\)
\(ABCD\)の体積\(V\)は、
\(\displaystyle V=\frac{1}{3}\times25\sqrt{3}\times\frac{10\sqrt{6}}{3}\)
\(\displaystyle \ \ =\frac{250\sqrt{2}}{3}\)cm\(^3\)

三平方の定理より、\(BD,BG,DG\)を求める。
\(BD^2=1^2+(\sqrt{6})^2=7\)
\(BD>0\)より、\(BD=\sqrt{7}\)
\(BG^2=(\sqrt{6})^2+(\sqrt{3})^2=9\)
\(BG>0\)より、\(BG=3\)
\(DG^2=1^2+(\sqrt{3})^2=4\)
\(DG>0\)より、\(DG=2\)
\(△BDG\)において、余弦定理より、
\(\cos∠BGD\)
\(\displaystyle =\frac{3^2+2^2-(\sqrt{7})^2}{2\times3\times2}\)
\(\displaystyle =\frac{6}{12}\)
\(\displaystyle =\frac{1}{2}\)
よって、
\(∠BGD=60^{\circ}\)

\(\displaystyle S=\frac{1}{2}\times2\times3\sin60^{\circ}\)
\(\displaystyle \ \ =3\times\frac{\sqrt{3}}{2}\)
\(\displaystyle \ \ =\frac{3\sqrt{3}}{2}\)

\(△BCD\)の底面積\(S\)は、
\(\displaystyle S=\frac{1}{2}\times1\times\sqrt{6}\)
\(\displaystyle \ \ =\frac{\sqrt{6}}{2}\)
\(BCDG\)の体積\(V\)は、
\(\displaystyle V=\frac{1}{3}\times\frac{\sqrt{6}}{2}\times\sqrt{3}\)
\(\displaystyle \ \ =\frac{3\sqrt{2}}{6}\)
\(\displaystyle \ \ =\frac{\sqrt{2}}{2}\)

\(\displaystyle V=\frac{1}{3}\times△BDG\times CI\)
\(\displaystyle CI=\frac{3V}{△BDG}\)
\(\displaystyle \ \ \ \ =3\times\frac{\sqrt{2}}{2}\div\frac{3\sqrt{3}}{2}\)
\(\displaystyle \ \ \ \ =\frac{\sqrt{2}}{\sqrt{3}}\)
\(\displaystyle \ \ \ \ =\frac{\sqrt{6}}{3}\)