【高校数学Ⅰ】3-1-1 鋭角の三角比｜問題集

\(\displaystyle \sin A=\frac{5}{13}\)

\(\displaystyle \cos A=\frac{12}{13}\)

\(\displaystyle \tan A=\frac{5}{12}\)

\(\displaystyle \sin B=\frac{1}{\sqrt{10}}\)

\(\displaystyle \cos B=\frac{3}{\sqrt{10}}\)

\(\displaystyle \tan B=\frac{1}{3}\)

三平方の定理より、斜辺の長さは\(\sqrt{5}\)
\(\displaystyle \sin C=\frac{1}{\sqrt{5}}\)

三平方の定理より、斜辺の長さは\(\sqrt{5}\)
\(\displaystyle \cos C=\frac{2}{\sqrt{5}}\)

\(\displaystyle \tan C=\frac{1}{2}\)

三平方の定理より、辺の長さは\(4\sqrt{2}\)
\(\displaystyle \sin D=\frac{4\sqrt{2}}{9}\)

\(\displaystyle \cos D=\frac{7}{9}\)

三平方の定理より、辺の長さは\(4\sqrt{2}\)
\(\displaystyle \tan D=\frac{4\sqrt{2}}{7}\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \frac{1}{\sqrt{2}}\)

\(\sqrt{3}\)

\(\displaystyle \frac{1}{\sqrt{2}}\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \frac{1}{\sqrt{3}}\)

\(\displaystyle \sin 19^{\circ}=\frac{x}{100}\)
\(x=100\sin 19^{\circ}\)
\(\ \ =100×0.3256\)
\(\ \ =32.56\)
よって、\(33\)m

\(\displaystyle \tan 21^{\circ}=\frac{x}{10}\)
\(x=10\tan 21^{\circ}\)
\(\ \ =10×0.3839\)
\(\ \ =3.839\)
よって、\(3.8+1.6=5.4\)m