Odpowiedź:
zadanie 5:
a = 10
b, c = ?
kąt = 45st.
tg(45) = [tex]\frac{a}{b} = \frac{10}{b}[/tex]
1 = [tex]\frac{10}{b}[/tex]
1b = 10
sin(45) = [tex]\frac{10}{c}[/tex]
[tex]\frac{\sqrt{2} }{2}[/tex] = [tex]\frac{10}{c}[/tex]
c = [tex]\frac{20}{\sqrt{2} }[/tex]
c = okolo 14.18cm
zadanie 6:
c = 20
kat: 30st
a, b = ?
sin(30) = [tex]\frac{a}{20}[/tex]
[tex]\frac{1}{2} = \frac{a}{20}\\\\a = \frac{20}{2}\\a = 10\\\\tg(30) = \frac{10}{b} \\\\\frac{\sqrt{3} }{2} = \frac{10}{b}\\\\b = \frac{2*10}{\sqrt{3} } = \frac{20}{\sqrt{3} } = 11.56[/tex]
zadanie 7:
alfa = 180 - 135 = 45st
tg(45) = [tex]\frac{1}{h}[/tex]
[tex]1 = \frac{1}{h} \\h = 1[/tex]
sin(45) = [tex]\frac{1}{a}[/tex]
[tex]\frac{\sqrt{2} }{2} = \frac{1}{a}\\\\a = \frac{2}{\sqrt{2} }[/tex]
Ob = 2a + 4 + 6 = 2 * [tex]\frac{2}{\sqrt{2} }[/tex] + 10 = [tex]\frac{4}{\sqrt{2} }[/tex] + 10 = okolo 12.83cm
P = [tex]\frac{(4+6) * h }{2} = \frac{10 * 1}{2} = \frac{10}{2} = 5cm^{2}[/tex]
Szczegółowe wyjaśnienie:
