Question

[Uyên Thảo]

1. Cho tứ giác ABCD có góc C=120, góc D=100, góc A-B=40. Tính góc A,B
2. Tính các góc tứ giác MNPQ biết
góc M:N:P:Q=1:3:4:7

Nhanh nha!

Answer 1

[Phạm Hoàng Hải Anh]

B1:Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

\(\Rightarrow\widehat{A}+\widehat{B}=360-\widehat{C}-\widehat{D}=360-120-100=140^0\)

Ta lại có : \(\widehat{A}-\widehat{B}=40^0\)

\(\Rightarrow\widehat{A}=40+\widehat{B}\)

Khi đó : \(\widehat{A}+\widehat{B}=140\)

\(\Rightarrow40+\widehat{B}+\widehat{B}=140\)

\(\Rightarrow2\widehat{B}=100^0\)

\(\Rightarrow\widehat{B}=50^0\)

\(\Rightarrow\)\(\widehat{A}=40+50=90^0\)

Answer 2

[Phạm Hoàng Hải Anh]

B2: Ta có : \(\widehat{M}:\widehat{N}:\widehat{P}:\widehat{Q}=1:3:4:7\)

\(\Rightarrow\frac{\widehat{M}}{1}=\frac{\widehat{N}}{3}=\frac{\widehat{P}}{4}=\frac{\widehat{Q}}{7}\)Và \(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^0\)

Áp dụng dãy tỉ số = nhau ta có :

\(\frac{\widehat{M}}{1}=\widehat{\frac{N}{3}}=\widehat{\frac{P}{4}}=\widehat{\frac{Q}{7}}=\frac{\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}}{1+3+4+7}=\frac{360}{15}=24\)

Khi đó : \(\widehat{M}=1.24=24^0\)

\(\widehat{N}=3.24=72^0\)

\(\widehat{P}=4.24=96^0\)

\(\widehat{Q}=24.7=168^0\)