Explicație pas cu pas:
teorema sinusurilor:
[tex]\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R \\[/tex]
1)
[tex]\frac{3}{\sin(30)} = 2R \\ [/tex]
[tex]R = \frac{3}{2 \times \frac{1}{2}} = \frac{3}{1} = > R = 3 \\ [/tex]
2)
[tex]\frac{8}{\sin(45)} = 2R \\ [/tex]
[tex]R = \frac{8}{2 \times \frac{ \sqrt{2} }{2} } = \frac{8}{ \sqrt{2} } = > R = 4 \sqrt{2} \\ [/tex]
3)
[tex]\frac{3}{\sin(A)} = 2 \times \frac{3}{2} \\ [/tex]
[tex]\frac{3}{\sin(A)} = 3 = > \sin(A) = 1 \\[/tex]
4)
[tex]\frac{2}{\sin(A)} = \frac{4}{\sin(60)}\\[/tex]
[tex]\frac{2}{\sin(A)} = \frac{4}{ \frac{ \sqrt{3} }{2} } = > \sin(A) = \frac{ \sqrt{3} }{4} \\[/tex]
5)
[tex]\frac{ \sqrt{2} }{\sin(30)} = \frac{AC}{\sin(45)} \\ [/tex]
[tex]AC = \frac{ \sqrt{2} \times \frac{ \sqrt{2} }{2} }{ \frac{1}{2} } = > AC = 2 \\ [/tex]
6)
m(<C) = 90° - 30° = 60°
[tex]\frac{BC}{\sin(90)} = \frac{4 \sqrt{3} }{\sin(60)} \\ [/tex]
[tex]\frac{BC}{1} = \frac{4 \sqrt{3} }{ \frac{ \sqrt{3} }{2} } = > BC = 8 \\ [/tex]
7)
[tex]\frac{20}{\sin(30)} = 2R \\ [/tex]
[tex]\frac{20}{ \frac{1}{2} } = 2R = > R = 20 \\ [/tex]
8)
[tex]\frac{AC}{\sin(45)} = \frac{10}{\sin(30)} \\[/tex]
[tex]AC = \frac{10 \times \frac{ \sqrt{2} }{2} }{ \frac{1}{2}} = > AC = 10 \sqrt{2} \\ [/tex]
9)
[tex]b = c[/tex]
[tex]\frac{b \times c}{2} = 18 = > bc = 36 \\ {b}^{2} = 36 = > b = 6 = > c = 6 [/tex]
10)
[tex]\frac{AB}{\sin(90)} = 2 \times 10 \\[/tex]
[tex] \frac{AB}{1} = 20 = > AB = 20 \\ [/tex]
