1.次の図形の面積\(S\)を求めなさい。
(1)\(b=3,c=2,A=60^{\circ}\)の\(△ABC\)
\(\displaystyle S=\frac{1}{2}\times3\times2\sin60^{\circ}\)
\(\displaystyle \ \ =3\times\frac{\sqrt{3}}{2}\)
\(\displaystyle \ \ =\frac{3\sqrt{3}}{2}\)
(2)\(a=7,b=5,c=6\)の\(△ABC\)
\(\displaystyle s=\frac{7+5+6}{2}=9\)
よって、
\(S=\sqrt{9(9-7)(9-5)(9-6)}\)
\(\ \ =\sqrt{216}\)
\(\ \ =6\sqrt{6}\)
(3)\(b=4,c=3,A=45^{\circ}\)の\(△ABC\)
\(\displaystyle S=\frac{1}{2}\times4\times3\sin45^{\circ}\)
\(\displaystyle \ \ =6\times\frac{1}{\sqrt{2}}\)
\(\displaystyle \ \ =3\sqrt{2}\)
(4)\(a=7,b=5,c=3\)の\(△ABC\)
\(\displaystyle s=\frac{7+5+3}{2}=\frac{15}{2}\)
よって、
\(\displaystyle S=\sqrt{\frac{15}{2}\left(\frac{15}{2}-7\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-3\right)}\)
\(\displaystyle \ \ =\sqrt{\frac{675}{16}}\)
\(\displaystyle \ \ =\frac{15\sqrt{3}}{4}\)
(5)\(b=10,c=8,A=45^{\circ}\)の\(△ABC\)
\(\displaystyle S=\frac{1}{2}\times10\times8\sin45^{\circ}\)
\(\displaystyle \ \ =40\times\frac{1}{\sqrt{2}}\)
\(\ \ =20\sqrt{2}\)
(6)\(a=2,b=2,A=30^{\circ}\)の\(△ABC\)
\(\displaystyle S=\frac{1}{2}\times2\times2\sin120^{\circ}\)
\(\displaystyle \ \ =2\times\frac{\sqrt{3}}{2}\)
\(\ \ =\sqrt{3}\)
(7)\(a=3,c=2,B=30^{\circ}\)の平行四辺形\(ABCD\)
\(\displaystyle S=2\times\frac{1}{2}\times3\times2\sin30^{\circ}\)
\(\displaystyle \ \ =6\times\frac{1}{2}\)
\(\ \ =3\)
2.\(△ABC\)において、\(∠A\)の二等分線と辺\(BC\)の交点を\(D\)とする。
(1)\(b=3,c=5,A=60^{\circ}\)のとき、辺\(AD\)を求めなさい。
\(AD=x\)とおく。
\(△ABC\)の面積を\(S\)とすると、
\(\displaystyle S=\frac{1}{2}\times3\times5\sin60^{\circ}\)
\(\displaystyle \ \ =\frac{15}{2}\times\frac{\sqrt{3}}{2}\)
\(\displaystyle \ \ =\frac{15\sqrt{3}}{4}\)
\(△ABD\)の面積を\(S_{1}\)とすると、
\(\displaystyle S_{1}=\frac{1}{2}\times5\times x\sin30^{\circ}\)
\(\displaystyle \ \ \ \ =\frac{5}{2}x\times\frac{1}{2}\)
\(\displaystyle \ \ \ \ =\frac{5}{4}x\)
\(△ACD\)の面積を\(S_{2}\)とすると、
\(\displaystyle S_{2}=\frac{1}{2}\times3\times x\sin30^{\circ}\)
\(\displaystyle \ \ \ \ =\frac{3}{2}x\times\frac{1}{2}\)
\(\displaystyle \ \ \ \ =\frac{3}{4}x\)
\(S=S_{1}+S_{2}\)より、
\(\displaystyle =\frac{15\sqrt{3}}{4}=\frac{5}{4}x+\frac{3}{4}x\)
\(\displaystyle x=\frac{15\sqrt{3}}{8}\)
よって、
\(\displaystyle AD=\frac{15\sqrt{3}}{8}\)
