Question

Exercitiile 5 si 6 - sunt de la subiectul 1 - Model Bacalaureat Mate Info
vreau cat mai explicit si cat mai detaliat . Ati putea pune poza cu rezolvarea . Dau si coronita .Vreau sa vad si formulele folosite si pasii nu doar rezolvarea.
Multumesc!

Answer 1

5)  Fie AM - mediana, unde M este mijlocul laturii BC.

$\it \left.\begin{aligned}\ \it x_M=\dfrac{x_B+x_C}{2}=\ \dfrac{2+6}{2}=4\\ \\ \\ \it y_M=\dfrac{y_B+y_C}{2}=\ \dfrac{-5+1}{2}=-2\end{aligned}\right\} \Rightarrow M(4,\ -2)$

$\it (AM):\ \dfrac{y-y_A}{y_M-y_A}=\dfrac{x-x_A}{x_M-x_A} \Rightarrow \dfrac{y-3}{-2-3}=\dfrac{x-1}{4-1} \Rightarrow\dfrac{y-3}{-5}=\dfrac{x-1}{3} \Rightarrow\\ \\ \\ \Rightarrow 3y-9=-5x+6|_{+9} \Rightarrow 3y=-5x+15|_{:3} \Rightarrow y=-\dfrac{5}{3}+5$

6)

$\it Vom\ aplica\ teorema\ sinisurilor:\\ \\ \dfrac{BC}{sinA}=2R \Rightarrow \dfrac{6}{sin\dfrac{\pi}{6}}=2R|_{:2} \Rightarrow R=\dfrac{3}{sin\dfrac{\pi}{6}}=\dfrac{3}{\dfrac{1}{2}}=3:\dfrac{1}{2}=3\cdot\dfrac{2}{1}=6$