Bài 1)
Ta có : A + B + C + D = 360 độ
=> A + B = 140 độ
Ta có :
A = \(\frac{140+40}{2}\)= 90 độ
=> B = 90 - 40 = 50 độ
Bài 1 :
Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
\(\Rightarrow\widehat{A}+\widehat{B}+120^o+100^o=360^o\)
\(\Rightarrow\widehat{A}+\widehat{B}+220^o=360^o\)
\(\Rightarrow\widehat{A}+\widehat{B}=140^o\)
Mà : \(\widehat{A}-\widehat{B}=40^o\)
\(\Rightarrow\widehat{A}+\widehat{A}+\widehat{B}-\widehat{B}=140^o+40^o\)
\(\Rightarrow2\widehat{A}=180^o\Leftrightarrow\widehat{A}=90^o\)
\(\Leftrightarrow\widehat{B}=140^o-\widehat{A}=140^o-90^o=50^o\)
\(KL:\widehat{A}=90^o;\widehat{B}=50^o\)
Bài 2)
Ta có M : N : P : Q = 1 : 3 : 4 : 7
=> M = N/3 = P/4 =Q/7
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
=> M + N+ P + Q /1 +3+4+7 = \(\frac{360}{15}\)=24
=> M = 24 độ
N = 72 độ
P = 96 độ
Q = 168 độ
1. Xét tứ giác ABCD có
\(\widehat{A}\)+ \(\widehat{B}\)+\(\widehat{C}\)+\(\widehat{D}\)= 360
\(\widehat{A}\)+\(\widehat{B}\)+ 120 + 100 = 360
\(\widehat{A}\)+\(\widehat{B}\) = 360 - 220
\(\widehat{A}\)+\(\widehat{B}\) = 140
\(\widehat{A}\)_ \(\widehat{B}\) = 40
\(\Rightarrow\)\(\widehat{A}\)= 140 + 40 / 2 = 90
\(\widehat{B}\)= 140 - 90 = 50
2. \(\frac{\widehat{M}}{1}\)=\(\frac{\widehat{N}}{3}\)=\(\frac{\widehat{P}}{4}\)=\(\frac{\widehat{Q}}{7}\)=\(\frac{\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}}{1+3+4+7}\)=\(\frac{360}{15}\)= 24
\(\frac{\widehat{M}}{1}\)= 36 \(\Rightarrow\)\(\widehat{M}\)= 24 * 1= 24
\(\frac{\widehat{N}}{3}\)= 24\(\Rightarrow\)\(\widehat{N}\)=24* 3 =72
\(\frac{\widehat{P}}{4}\)= 24\(\Rightarrow\)\(\widehat{P}\)= 24 * 4 = 96
\(\frac{\widehat{Q}}{7}\)= 24 \(\Rightarrow\)\(\widehat{Q}\)= 24* 7 = 168
