Question

Va rog ajutati-ma, nu stiu cum sa fac!!

Answer 1

[tex]\it cosB=\dfrac{3}{5} tgC\ \ \ \ (*) \\ \\ \\ cosB=\dfrac{AB}{BC}=\dfrac{4\sqrt2}{BC} \\ \\ \\ tgC= \dfrac{AB}{AC} =\dfrac{4\sqrt2}{AC}[/tex]

Relația din enunț devine:

[tex]\it \dfrac{4\sqrt2}{BC} =\dfrac{3}{5}\cdot \dfrac{4\sqrt2}{AC} |_{\cdot\frac{AC}{4\sqrt2} } \Rightarrow \dfrac{AC}{BC} =\dfrac{3}{5} \Rightarrow \left(\dfrac{AC}{BC} \right)^2 =\left(\dfrac{3}{5}\right)^2 \Rightarrow \\ \\ \\ \Rightarrow \dfrac{AC^2}{BC^2} =\dfrac{9}{25 } \Rightarrow AC^2= \dfrac{9\cdot BC^2}{25} \ \ \ \ (1)[/tex]

Din teoreme lui Pitagora, rezultă:

[tex]\it AC^2= BC^2-AB^2 = BC^2-(4\sqrt2)^2 =BC^2-32\ \ \ \ (2) \\ \\ \\ (1), (2) \Rightarrow BC^2-32=\dfrac{9 BC^2}{25} \Rightarrow 25BC^2-32\cdot25 =9BC^2 \Rightarrow \\ \\ \\ \Rightarrow 25BC^2-9BC^2=32\cdot25 \Rightarrow 16BC^2=32\cdot25|_{:16} \Rightarrow BC^2=2\cdot 25[/tex]

$\it \Rightarrow BC=\sqrt{2\cdot25} \Rightarrow BC= 5\sqrt2\ cm$

Answer 2

ai Pb rezolvata......