pole stożka
pole stożka
Wiedząc, że pole powierzchni całkowitej stożka wynosi 360 cm kw., a pole powierzchni bocznej wynosi 240 cm kw., wyznacz tg kąta nachylenia tworzącej do płaszczyzny podstawy.?!?!?!?!
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barakuda
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- Rejestracja: 22 paź 2009, o 19:37
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pole stożka
\(\displaystyle{ P_{c}=P_{p}+P_{b}}\)
\(\displaystyle{ 360 = \pi \cdot r^2 +240}\)
\(\displaystyle{ r^2 = \frac{120}{\pi}}\)
\(\displaystyle{ r = \frac{2 \sqrt{30} }{ \sqrt{\pi} } = \frac{2 \sqrt{30 \pi} }{\pi}}\)
\(\displaystyle{ P_{b}= \pi \cdot r \cdot l = 240}\)
\(\displaystyle{ \pi \cdot \frac{2 \sqrt{30 \pi} }{\pi} \cdot l =240}\)
\(\displaystyle{ l= \frac{240}{2 \sqrt{30 \pi} } = \frac{4 \sqrt{30 \pi} }{\pi}}\)
\(\displaystyle{ h= \sqrt{l^2 - r^2} = \sqrt{(\frac{4 \sqrt{30 \pi} }{\pi})^2 -(\frac{2 \sqrt{30 \pi} }{\pi} )^2} = \sqrt{ \frac{480}{\pi}- \frac{120}{\pi} } = \sqrt{ \frac{360}{\pi} } = \frac{6 \sqrt{10 \pi} }{\pi}}\)
\(\displaystyle{ tg\alpha = \frac{h}{r} = \frac{\frac{6 \sqrt{10 \pi} }{\pi} }{\frac{2 \sqrt{30 \pi} }{\pi} } = \frac{6 \sqrt{10 \pi} }{\pi} \cdot \frac{\pi}{2 \sqrt{30 \pi} } = \frac{3}{ \sqrt{3} } = \sqrt{3} = 60^o}\)
\(\displaystyle{ 360 = \pi \cdot r^2 +240}\)
\(\displaystyle{ r^2 = \frac{120}{\pi}}\)
\(\displaystyle{ r = \frac{2 \sqrt{30} }{ \sqrt{\pi} } = \frac{2 \sqrt{30 \pi} }{\pi}}\)
\(\displaystyle{ P_{b}= \pi \cdot r \cdot l = 240}\)
\(\displaystyle{ \pi \cdot \frac{2 \sqrt{30 \pi} }{\pi} \cdot l =240}\)
\(\displaystyle{ l= \frac{240}{2 \sqrt{30 \pi} } = \frac{4 \sqrt{30 \pi} }{\pi}}\)
\(\displaystyle{ h= \sqrt{l^2 - r^2} = \sqrt{(\frac{4 \sqrt{30 \pi} }{\pi})^2 -(\frac{2 \sqrt{30 \pi} }{\pi} )^2} = \sqrt{ \frac{480}{\pi}- \frac{120}{\pi} } = \sqrt{ \frac{360}{\pi} } = \frac{6 \sqrt{10 \pi} }{\pi}}\)
\(\displaystyle{ tg\alpha = \frac{h}{r} = \frac{\frac{6 \sqrt{10 \pi} }{\pi} }{\frac{2 \sqrt{30 \pi} }{\pi} } = \frac{6 \sqrt{10 \pi} }{\pi} \cdot \frac{\pi}{2 \sqrt{30 \pi} } = \frac{3}{ \sqrt{3} } = \sqrt{3} = 60^o}\)