$\textbf{Bài 4:}$
$\text{a/ Ta có: AB ⊥ p (giả thiết)}$
$\text{p // q (gt)}$
$\text{nên AB ⊥ q (T/c 2 quan hệ từ vuông góc đến song song)}$
$\text{Vậy AB ⊥ q}$
$\text{b/ Ta có: $\widehat{D_1}$ = $\widehat{D_2}$ = $70^o$ (đối đỉnh)}$
$\text{Vậy $\widehat{D_2}$ = $70^o$}$
$\text{c/ AB ⊥ q (cm ở câu a)}$
$\text{nên $\widehat{B_1}$ = $90^o$}$
$\text{p // q (gt)}$
$\text{nên $\widehat{D_1}$ + $\widehat{C_2}$ = $180^o$ (tcp)}$
$\text{T/s: $70^o$ + $\widehat{C_2}$ = $180^o$}$
$\text{⇒ $\widehat{C_2}$ = $180^o$ - $70^o$ = $110^o$}$
$\text{Vậy $\widehat{B_1}$ = $90^o$, $\widehat{C_2}$ = $110^o$}$
$\text{HỌC TỐT!}$
$Hry_{123}$
