Question

5 si 6 va rog
...multumesc

Answer 1

La 5) sunt mai multe metode de rezolvare, eu iti dau rezolvarea in metoda analitica
A-B-C ⇔ $m_{AB}=m_{BC}$ (panta lui AB = panta lui BC)

$m_{AB}= \frac{y_{B}- y_{A} }{x_{B} - x_{A}} \\ \\ m_{AB}= \frac{2-1}{3-1} \\ \\ m _{AB}= \frac{1}{2}$


$m_{BC}= \frac{y_{C} - y_{B} }{ x_{C} - x_{B} } \\ \\ m_{BC}= \frac{3-2}{m-3} \\ \\ m_{BC}= \frac{1}{m-3}$


$m_{AB} = m_{BC} \\ \\ \frac{1}{2}= \frac{1}{m-3} \\ \\ m-3=2 \\ \\ m=5$


6) sin²70° + sin²20° =
 = [sin(90°-20°)]² + sin²20° =
 = (sin 90° * cos 20° - cos 90° * sin 20°)² + sin²20° =
 = (1*cos 20° - 0*sin 20°)² + sin²20° =
 = (cos 20°)² + sin²20° =
 = sin²20° + cos²20° =
 = 1