daje naj I podziękowania:)​

Daje Naj I Podziękowania class=

Odpowiedź :

a)

[tex] \beta = {30}^{o} [/tex]

[tex]b = 2 \sqrt{3} \times \sqrt{3} = 2 \sqrt{9} = 2 \times 3 = 6[/tex]

[tex]c = 4 \sqrt{3} [/tex]

b)

[tex] \beta = {60}^{o} [/tex]

[tex]a = 2 \sqrt{2} [/tex]

[tex]b = 2 \sqrt{2} \times \sqrt{3} = 2 \sqrt{6} [/tex]

c)

[tex] \beta = {30}^{o} [/tex]

[tex]a = \frac{9}{ \sqrt{3} } = \frac{9 \sqrt{3} }{3} = 3 \sqrt{3} [/tex]

[tex]c = 6 \sqrt{3} [/tex]

d)

[tex] \beta = {60}^{o}[/tex]

[tex]a = \frac{ \sqrt{6} }{2} \div 2 = \frac{ \sqrt{6} }{2} \times \frac{1}{2} = \frac{ \sqrt{6} }{4} [/tex]

[tex]b = \frac{ \sqrt{6} }{4} \times \sqrt{3} = \frac{ \sqrt{18} }{4} = \frac{ \sqrt{9 \times 2} }{4} = \frac{3 \sqrt{2} }{4} [/tex]

e)

[tex] \beta = {30}^{o} [/tex]

[tex]a = \frac{ \sqrt{5} }{ \sqrt{3} } = \frac{ \sqrt{15} }{3}[/tex]

[tex]c = 2 \times \frac{ \sqrt{15} }{3} = \frac{2 \sqrt{15} }{3} [/tex]

f)

[tex] \beta = {60}^{o} [/tex]

[tex]b = \sqrt{12} \times \sqrt{3} = \sqrt{36} = 6[/tex]

[tex]c = 2 \sqrt{12} = 2 \sqrt{4 \times 3} = 2 \times 2 \sqrt{3} = 4 \sqrt{3} [/tex]