Bardzo was Proszę o szybką pomoc! cała strona

Odpowiedź: proszę bardzo!
Szczegółowe wyjaśnienie:
1. a
[tex] \beta = 60[/tex]
[tex]6 = x \sqrt{3} \\ c = 2b \\ 6 = b \sqrt{3} \\ b = \frac{6}{ \sqrt{3} } = \frac{6 \sqrt{3} }{3} = 2 \sqrt{3} \\ c = 2 \times 2 \sqrt{3} = 4 \sqrt{3} [/tex]
1 b.
[tex] \alpha = 30 \\ c = 2a \\ 4 = a \sqrt{3} \\ a = \frac{4}{ \sqrt{3} } = \frac{4 \sqrt{3} }{3} \\ c = 2 \times \frac{4 \sqrt{3} }{3} = \frac{8 \sqrt{3} }{3 } [/tex]
2.
a = 10cm
b = 24cm
[tex]{c}^{2} = {10}^{2} + {24}^{2} \\ {c}^{2} = 100 + 576 \\ c = \sqrt{676} = 26[/tex]
[tex] \tan( \alpha ) = \frac{10}{24} = \frac{5}{14} = 0.3571 \\ \alpha = 20 \\ \beta = 180 - 90 - 20 = 180 - 110 = 70[/tex]
Trójkąt o bokach 10, 24 i 26cm oraz kątach 90, 70, 20 stopni.
3.
[tex] \alpha = 40 \\ a = 6cm[/tex]
[tex] \sin(40) = \frac{6}{c} \\ \frac{6428}{10000} = \frac{6}{c} \\ c = \frac{60000}{6428} = 9.3cm \\ {9.3}^{2} = {6}^{2} + {b}^{2} \\ {b}^{2} = 86.49 - 36 \\ b = \sqrt{50.49} = 7.1 \\ \beta = 180 - 90 - 40 = 180 - 130 = 50[/tex]
Trójkąt o bokach 6cm, 7.1cm i 9.3cm oraz kątach 90, 50, 40