Oznaczmy boki trójkąta przez a,b,c.
\tan\alpha=\frac{b}{a}=4
b=4a
\cot =\frac{1}{\tan}=\frac{1}{4}
c^2=a^^2+b^2
c^2=a^2+(4a)^2
c^2=a^2+16a^2
c^2=a^2(1+16)
c=a\sqrt{1+16}
\sin=\frac{b}{c}=\frac{4a}{a\sqrt{1+16}}*\frac{\sqrt{1+16} }{\sqrt{1+16}}=\frac{4\sqrt{1+16}}{1+16}=\frac{4}{17}*\sqrt{1+16}
\cos=\frac{a}{c}=\frac{a}{a\sqrt{1+16}}*\frac{\sqrt{1+16} }{\sqrt{1+16}}=\frac{1}{17}*\sqrt{1+16}