Question

Va rog explicati-mi rezolvarea acestei probleme. In special de unde au scos AC=3c rad.2/4.
Multumesc!

Answer 1

Se aplica Pitagora și se tine seama de faptul că AC=3BC

Answer 2

[tex]\it sinA = \dfrac{1}{3} \ \ \ \ \ \ (1) \\ \\ \\ sinA= \dfrac{BC}{AC} \ \ \ \ \ (2) \\ \\ \\ (1),\ (2) \Rightarrow \dfrac{BC}{AC} = \dfrac{1}{3} \Rightarrow AC=3BC\ \ \ \ (3)[/tex]

[tex]\it Notez\ BC = x,\ iar\ din\ (3) \Rightarrow AC = 3x \\ \\ \\ Cu \ teorema \ lui\ Pitagora \Rightarrow AC^2=AB^2+BC^2 \Rightarrow (3x)^2= c^2+x^2 \Rightarrow \\ \\ \\ 9x^2= c^2+x^2 \Rightarrow 8x^2=c^2 \Rightarrow x^2= \dfrac{c^2}{8} \Rightarrow x^2=\dfrac{c^2}{(2\sqrt2)^2} \Rightarrow x = \dfrac{c}{2\sqrt2} [/tex]

[tex]\it \Rightarrow x = \dfrac{c\sqrt2}{2\cdot2} \Rightarrow x = \dfrac{c\sqrt2}{4} \Rightarrow \begin{cases} \it BC = \dfrac{c\sqrt2}{4} \\ \\ \it AC = 3\cdot\dfrac{c\sqrt2}{4} = \dfrac{3c\sqrt2}{4}\end{cases}[/tex]