\(a//b;a\perp AB\Rightarrow b\perp AB\Rightarrow\widehat{B}=90^0\)
\(a//b\Rightarrow\widehat{D}+\widehat{C}=180^0\left(trong.cùng.phía\right)\\ \Rightarrow\widehat{D}=180^0-130^0=50^0\)
a,Ta thấy \(\widehat{B}\) là góc vuông
\(\Rightarrow\widehat{B}=90^o\)
\(\widehat{C}+\widehat{D}=180^o\)(bù nhau)
\(\Rightarrow\widehat{D}=180^o-130^o=50^o\)
Vì a//b (gt)
=> góc A = góc B(slt)
Mà góc A = 90°(gt)
=> góc B = 90°
Vì a//b(gt)
=> góc C + góc D = 180°(tcp)
=> 130°+ góc D = 180°
góc D = 180°- 130° = 50°
Ta có: a// b ⇒ \(\widehat{A}\)= \(\widehat{B}\)( 2 góc đồng vị )
⇒\(\widehat{B}\)= 90'
Ta lại có: a//b ⇒ \(\widehat{C}\) + \(\widehat{D}\)= 180'( 2 góc trong cùng phía)
⇒130'+\(\widehat{D}\)= 180'
⇒\(\widehat{D}\)= 50'
Do a//b ⇒ \(\widehat{A}+\widehat{B}=180^o\left(2gócTCP\right)\)
\(\Rightarrow90^o+\widehat{B}=180^o\)
\(\Rightarrow\widehat{B}=180^o-90^o=90^o\)
Do a//b \(\Rightarrow\widehat{D}+\widehat{C}=180^o\left(2gócTCP\right)\)
\(\Rightarrow\widehat{D}+130^o=180^o\)
\(\Rightarrow\widehat{D}=180^o-130^0=50^o\)