Question

Primul exercitiu va rog

Answer 1

,ai rezolvarea în imagine

Answer 2

a)

[tex]\it \ Ducem \ \ AA' \perp\alpha,\ \ A' \in\alpha \Rightarrow AA' = d(A,\ \alpha) =3\ cm \\\;\\ \\\;\\ B\in\alpha,\ \ AA'\perp\alpha \Rightarrow pr_{\alpha} [AB] = [A'B] \\\;\\ \\\;\\ \Delta A'AB-dr\ \stackrel{T.P.}{\Longrightarrow} A'B^2 = AB^2- A'B^2 = 6^2-3^2= \\\;\\ \\\;\\ =(6-3)(6+3) =3\cdot9 \Rightarrow A'B = \sqrt{3\cdot9} = 3\sqrt3\ cm [/tex]

b)

[tex]\it m(\widehat{AB,\ \alpha}) = m(\widehat{AB,\ A'B}) = m(\hat{B}) \\\;\\ AA' = 3 = \dfrac{6}{2} =\dfrac{AB}{2} \stackrel{R.T. \angle\ 30^o}{\Longrightarrow} m(\hat{B}) = 30^o \Rightarrow m(\widehat{AB,\ \alpha}) = 30^o[/tex]