Question

Cine știe, va rog, dau coroana urgent. Rezolvarea toată sau ce știți voi ​

Answer 1

Explicație pas cu pas:

teorema sinusurilor:

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$

1)

$\frac{3}{\sin(30)} = 2R$

$R = \frac{3}{2 \times \frac{1}{2}} = \frac{3}{1} = > R = 3$

2)

$\frac{8}{\sin(45)} = 2R$

$R = \frac{8}{2 \times \frac{ \sqrt{2} }{2} } = \frac{8}{ \sqrt{2} } = > R = 4 \sqrt{2}$

3)

$\frac{3}{\sin(A)} = 2 \times \frac{3}{2}$

$\frac{3}{\sin(A)} = 3 = > \sin(A) = 1$

4)

$\frac{2}{\sin(A)} = \frac{4}{\sin(60)}$

$\frac{2}{\sin(A)} = \frac{4}{ \frac{ \sqrt{3} }{2} } = > \sin(A) = \frac{ \sqrt{3} }{4}$

5)

$\frac{ \sqrt{2} }{\sin(30)} = \frac{AC}{\sin(45)}$

$AC = \frac{ \sqrt{2} \times \frac{ \sqrt{2} }{2} }{ \frac{1}{2} } = > AC = 2$

6)

m(<C) = 90° - 30° = 60°

$\frac{BC}{\sin(90)} = \frac{4 \sqrt{3} }{\sin(60)}$

$\frac{BC}{1} = \frac{4 \sqrt{3} }{ \frac{ \sqrt{3} }{2} } = > BC = 8$

7)

$\frac{20}{\sin(30)} = 2R$

$\frac{20}{ \frac{1}{2} } = 2R = > R = 20$

8)

$\frac{AC}{\sin(45)} = \frac{10}{\sin(30)}$

$AC = \frac{10 \times \frac{ \sqrt{2} }{2} }{ \frac{1}{2}} = > AC = 10 \sqrt{2}$

9)

$b = c$

$\frac{b \times c}{2} = 18 = > bc = 36 \\ {b}^{2} = 36 = > b = 6 = > c = 6$

10)

$\frac{AB}{\sin(90)} = 2 \times 10$

$\frac{AB}{1} = 20 = > AB = 20$