Question

Cho hình thang ABCD, AB // CD, \(\widehat{A}\) = 2\(\widehat{C}\), \(\widehat{B}-\widehat{C}=40^o\). Tính các góc hình thang.

Answer 1

vì AB // CD

⇒ \(\widehat{B}+\widehat{C}=180^O\) (2 góc trong cùng phía)

mà \(\widehat{B}-\widehat{C}=40^o\)

\(\Rightarrow\widehat{B}=\left(180^o+40^o\right):2=110^o\)

\(\Rightarrow\widehat{C}=110^o-40^o=70^o\)

\(\Rightarrow\widehat{A}=2.70^o=140^o\)

vì AB // CD

\(\Rightarrow\widehat{A}+\widehat{D}=180^o\) (2 góc trong cùng phía)

hay \(140^o+\widehat{D}=180^O\)

\(\Rightarrow\widehat{D}=40^o\)

vậy \(\widehat{A}=140^o,\widehat{B}=110^o,\widehat{C}=70^o,\widehat{D}=40^o\)

Answer 2

Góc B-Góc C=40

Góc B+Góc C=180(trong cùng phía)

Góc C=(180-40):2=70

Góc B=(180+40):2=110

Góc A=70.2=140

Góc D= 360-140-110-70=40