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[decha21]

Bardzo proszę o rozwiązanie zadań 5.2 i 5.3

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[eresh]

5,2a)
\(|AO|=\frac{1}{2}8\sqrt{2}\sqrt{2}\\
|AO|=\frac{1}{2}\cdot 8\cdot 2=8\\
\tg\alpha=\frac{|OS|}{|OA|}\\
\tg\alpha=\frac{8}{8}\\
\tg\alpha=1\\
\alpha=45^{\circ}\)

[eresh]

5.2b)
E - środek boku BC
\(|SO|^2+|OE|^2=|SE|^2\\
(3\sqrt{2})^2+3^2=|SE|^2\\
27=|SE|^2\\
|SE|=3\sqrt{3}\\
\tg\alpha=\frac{|SE|}{|EB|}\\
\tg\alpha=\frac{3\sqrt{3}}{3}\\
\tg\alpha=\sqrt{3}\\
\alpha=60^{\circ}\)

[eresh]

5.2c)
E - środek BC
\(\sin(\frac{1}{2}\alpha)=\frac{|EC|}{|CS|}\\
\sin\frac{\alpha }{2}=\frac{6}{4\sqrt{3}}\\
\sin\frac{\alpha}{2}=\frac{\sqrt{3}}{2}\\
0,5\alpha=60^{\circ}\\
\alpha = 120^{\circ}\)

[eresh]

5.3a)
\(|OC|=\frac{1}{2}|AC|=\frac{1}{2}\cdot \frac{2}{\sqrt{2}-1}\cdot\sqrt{2}=2+\sqrt{2}\\
\tg 45^{\circ}=\frac{x}{|OC|}\\
1=\frac{x}{2+\sqrt{2}}\\
x=2+\sqrt{2}\)

[eresh]

5.3b)
\(x=|DS|=|SA|\\
|AO|=4\sqrt{3}\\
\cos 30^{\circ}=\frac{|AO|}{|SA|}\\
\frac{\sqrt{3}}{2}=\frac{4\sqrt{3}}{x}\\
x\sqrt{3}=8\sqrt{3}\\
x=8\)

[eresh]

53.c)
D - środek krawędzi AB
\(|DO|=\frac{1}{3}|CD|=\frac{1}{3}\cdot\frac{x\sqrt{3}}{2}=\frac{x\sqrt{3}}{6}\\
\tg 60^{\circ}=\frac{|SO|}{|DO|}\\
\sqrt{3}=\frac{20}{\frac{x\sqrt{3}}{6}}\\
\frac{3x}{6}=20\\
\frac{x}{2}=20\\
x=40\)