Question

ajutor nu știu la ex 6​

Answer 1

Răspuns:

<A+<B+<C=180°

a) A+A+2×A=180

4A=180

A=45°

B=A=45°

C=2×A=2×45=90°

b)A=2×B ,C=3×B+12°

A+B+C=180°

2B+B+3B+12=180°

6B=180-12

6B=168

B=168/6

B=28°

A=2×28

A=56°

C=3×28+12

C=96°

c)A=3×C=>A/3=C ,B=A+12

A+B+C=180°

A+A+12+A/3=180

2A+A/3=180-12

2A+A/3=168

7A/3=168

7A=158×3

7A=504

A=72°

B=72+12

B=84°

72/3=C

C=24°

am notat A unghiul A ,B ungiul B si C unghiul C

Answer 2

$\it a)\\ \\ \hat B=\hat A\ \ \ \ \ \ (1)\\ \\ \hat C=2\hat A\ \ \ \ (2)\\ \\ \hat A+\hat B+\hat C=180^o\ \stackrel{(1),(2)}{\Longrightarrow}\ \hat A+\hat A+2\hat A=180^o \Rightarrow \\ \\ \Rightarrow4\hat A=180^o|_{:4} \Rightarrow \hat A=45^o\ \ \ \ \ (3)$

$\it (1),\ (3) \Rightarrow \hat B=45^o\\ \\ (2),\ (3) \Rightarrow \hat C=2\cdot45^o=90^o$

$\it b)\\ \\ \hat A=2\hat B\ \ \ \ \ \ (1)\\ \\ \hat C=3\hat B+12^o\ \ \ \ (2)\\ \\ \hat A+\hat B+\hat C=180^o\ \stackrel{(1),(2)}{\Longrightarrow}\ 2\hat B+\hat B+3\hat B+12^o=180^o|_{-12^o} \Rightarrow \\ \\ \Rightarrow 6\hat B=168^o|_{:6} \Rightarrow \hat B=28^o\ \ \ \ \ (3)$

$\it (1),\ (3) \Rightarrow \hat A=2\cdot28^o=56^o\\ \\ (2),\ (3) \Rightarrow \hat C=3\cdot28^o+12^o =96^o$

$\it c)\\ \\ \hat A=3\hat C\ \ \ \ \ \ (1)\\ \\ \hat B=\hat A+12^o=3\hat C+12^o\ \ \ \ (2)\\ \\ \hat A+\hat B+\hat C=180^o\ \stackrel{(1),(2)}{\Longrightarrow}\ 3\hat C+3\hat C+12^o+\hat C=180^o|_{-12^o} \Rightarrow \\ \\ \Rightarrow 7\hat C=168^o|_{:7} \Rightarrow \hat C=24^o\ \ \ \ \ (3)$

$\it (1),\ (3) \Rightarrow \hat A=3\cdot24^o=72^o\\ \\ (2),\ (3) \Rightarrow \hat B=3\cdot24^o+12^o =84^o$