5.
Z zależności boków w trójkącie prostokątnym o kątach ostrych 30° i 60° mamy:
[tex]a = 3\sqrt{2}\\\\c = 2a = 2\cdot3\sqrt{a}\\\\\boxed{c = 6\sqrt{2}}[/tex]
6.
[tex]\frac{\frac{a}{2}}{12} = cos45^{o}\\\\\frac{a}{24} = \frac{\sqrt{2}}{2}\\\\2a = 24\sqrt{2} \ \ /:2\\\\\boxed{a = 12\sqrt{2}}[/tex]
7.
[tex]cos\alpha = \frac{1}{4}\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1 - cos^{2}\alpha = 1-(\frac{1}{4})^{2} = \frac{16}{16}-\frac{1}{16}=\frac{15}{16}\\\\sin\alpha = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{\sqrt{16}}\\\\\boxed{sin\alpha = \frac{\sqrt{15}}{4}}[/tex]
[tex]tg\alpha = \frac{sin\alpha}{cos\alpha}\\\\tg\alpha = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}}\\\\\boxed{tg\alpha = \frac{\sqrt{15}}{15}}[/tex]
8.
[tex]tg\alpha = 2\\\\\frac{sin\alpha}{cos\alpha} = 2\\\\sin\alpha = 2cos\alpha\\\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\(2cos\alpha)^{2}+cos^{2}\alpha = 1\\\\4cos^{2}\alpha+cos^{2}\alpha = 1\\\\5cos^{2}\alpha = 1 \ \ /:5\\\\cos^{2}\alpha = \frac{1}{5}\\\\cos\alpha = \sqrt{\frac{1}{5}}=\frac{\sqrt{1}}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}\\\\\boxed{cos\alpha = \frac{\sqrt{5}}{5}}[/tex]
[tex]sin\alpha = 2\cdot\frac{\sqrt{5}}{5}\\\\\boxed{sin\alpha = \frac{2\sqrt{5}}{5}}[/tex]