Thursday, April 16, 2020

Right Triangle



\( α+ß=\frac{π}2=90°\ ,\ R=\frac{C}2 \)

\( r=\sqrt{{\left({\frac{C}2}\right)}^2+\frac{A·B}2}-\frac{C}2←\cases{A=r+x\\ B=r+y\\ C=x+y} \)

\( \frac{a}h=\frac{h}b=\frac{A}B=tan\ ß=\frac1{tan\ α}\)

\( a·b=h^2 \)

\( a+b=C \)

\( A^2+B^2=C^2={\left({a+b}\right)}^2=a^2+2·a·b+b^2 \)

\( A^2=a^2+h^2=a^2+a·b=a·\left({a+b}\right)=a·C \)

\( B^2=b^2+h^2=b^2+b·a=b·\left({b+a}\right)=b·C \)

\( a=\frac{A^2}C \)

\( b=\frac{B^2}C \)

\( h=\frac{A·B}C=\sqrt{a·b} \)

 \( S=\frac{A·B}2=\frac{\sqrt{a·C·b·C}}2=\frac{h·C}2=\frac{\frac{A·B}C·C}2 \)

 \( P=A+B+C=A+B+\sqrt{A^2+B^2}=\sqrt{a·C}+\sqrt{b·C}+\sqrt{C·C}=\sqrt{C}·\left({\sqrt{a}+\sqrt{b}+\sqrt{a+b}}\right) \)

\( b=h^2\ ,\ a=1\ →\ A^2=C=h^2+1\ ,\ B^2=b·A^2=b·C=h^2·\left({h^2+1}\right)\ ,\ R=\frac{C}2 \)


[Eop]