Matematyka.pl

Piotrek89 pisze:W celu znalezienia wierzchołków prostokąta rozwiązuje układ równań:

\(\displaystyle{ \begin{cases} x^{2}+2y^{2}=24\\x^{2}-2y^{2}=8\end{cases}}\)

Dodaję stronami:
\(\displaystyle{ 2x^{2}=32}\)
\(\displaystyle{ x^{2}=16}\)
\(\displaystyle{ x=4 x=-4}\)

1.dla x=4 :
\(\displaystyle{ 16+2y^{2}=24}\)
\(\displaystyle{ y^{2}=4}\)
\(\displaystyle{ y=2 y=-2}\)

\(\displaystyle{ \begin{cases} x=4\\y=2\end{cases} \begin{cases} x=4\\y=-2\end{cases}}\)

2.dla x=-4
\(\displaystyle{ 16+2y^{2}=24}\)
\(\displaystyle{ y^{2}=4}\)
\(\displaystyle{ y=2 y=-2}\)

\(\displaystyle{ \begin{cases} x=-4\\y=2\end{cases} \begin{cases} x=-4\\y=-2\end{cases}}\)

A,B,C,D - wierzchołki prostokąta:

A(-4,-2) B(4,-2) C(4,2) D(-4,2)

img518.imageshack. us/my.php?image=skanuj0002ly8.jpg

\(\displaystyle{ a=|SC|=\sqrt{4^{2}+2^{2}}=\sqrt{20}}\)
\(\displaystyle{ x=|BC|=4}\)

Z twierdzenia cosinusów:
\(\displaystyle{ x^{2}=a^{2}+a^{2}-2a^{2}\cdot \cos\alpha}\)
\(\displaystyle{ 16=20+20-2\cdot20\cdot \cos\alpha}\)
\(\displaystyle{ 40\cos\alpha=24}\)
\(\displaystyle{ \cos\alpha=\frac{24}{40}}\)
Odp: \(\displaystyle{ \cos\alpha=\frac{3}{5}}\)