Odpowiedź:
Szczegółowe wyjaśnienie:
h = 8cm
d - średnica podstawy
r - promień podstawy
p - przekątna walca
a)
Jeżeli przekątna p jest nachylona pod kątem a = 30° do podstawy walca, to z funkcji trygonometrycznych:
[tex]sin a = \frac{p}{h}[/tex]
cot 30° = [tex]\sqrt{3}[/tex]
[tex]cot a = \frac{d}{h} \\\sqrt{3} = \frac{d}{8cm} \\\\d = 8\sqrt{3} cm \\[/tex]
[tex]r = 4\sqrt{3}[/tex]
[tex]Pc = Pb + 2Pp\\Pp = \pi r^{2} = \pi * (4\sqrt{3}) ^{2} = 48\pi = 150,72cm^{2}\\Pb = \pi d * h= 8\sqrt{3} \pi * 8cm= 348.08cm^{2}\\Pc = 348.08 + 2*150,72 = 649.52cm^{2}[/tex]
b) Jeżeli sina = [tex]\sqrt{\frac{3}{2} }[/tex] to:
[tex]\sqrt{\frac{3}{2} } = \frac{h}{p} \\\\\sqrt{\frac{3}{2} } = \frac{8}{p} \\\\p = 8 \sqrt{\frac{2}{3} } \\\\\\d^{2} = p^{2} - h^{2}[/tex]
[tex]d^{2} = (8\sqrt{\frac{2}{3} } )^{2} * 8^{2} = \frac{64}{3} \\d = \sqrt{\frac{64}{3} } = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3} }{3} \\r = \frac{8\sqrt{3} }{6} \\\\Pp = \pi r^{2} = 16.76cm^{2} \\Pb = \pi dh = 116.08cm^{2}\\Pc = 2Pp + Pb = 149.6cm^{2}[/tex]
c) Jeżeli cos a = [tex]\frac{1}{4}[/tex] to:
[tex]cos a= \frac{d}{p} \\\\\frac{1}{4} = \frac{d}{p}[/tex]
[tex]sin^{2} a + cos^{2}a = 1\\sin^{2}a = 1 - cos^{2}a\\cos^{2}a = (\frac{1}{4} )^{2} = \frac{1}{16} \\sin^{2}a = 1 - \frac{1}{16} = \frac{15}{16} \\sina = \sqrt{\frac{15}{16} } = \frac{\sqrt{15} }{4} \\sin a = \frac{h}{d} \\d = h * sin a = 8 * \frac{\sqrt{15} }{4} = 2\sqrt{15} \\r = \sqrt{15} \\Pp = \pi r^{2} = 47.12cm^{2}\\Pb = \pi d *h= \pi * 2\sqrt{15} * 8 = 194.68cm^{2}\\Pc = 2Pp + Pb = 288.92cm^{2}[/tex]
