Question

I=70 N=60 Q=30 Tính M1,M2, NMQ, I2

Answer 1

xét ΔMNI có

\(\widehat{M1}+\widehat{N}+\widehat{I1}=180^0\) (3 góc trong 1 tam giác)

\(\rightarrow\widehat{M1}=180-\widehat{N}-\widehat{I1}=180-60-70\\ \rightarrow\widehat{M1}=50^0\)

lại có

\(\widehat{I1}+\widehat{I2}=180^0\left(kề-bù\right)\\ \rightarrow\widehat{I2}=180-\widehat{I1}=180-70\\ \rightarrow\widehat{I2}=110^0\)

xét ΔMQI có

\(\widehat{M2}+\widehat{Q}+\widehat{I2}=180^0\) (3 góc trong 1 tam giác)

\(\rightarrow\widehat{M2}=180-\widehat{Q}-\widehat{I2}=180-30-110\\ \rightarrow\widehat{M2}=40^0\)

mà 

\(\widehat{M1}+\widehat{M2}=\widehat{NMQ}\\ \rightarrow\widehat{MNQ}=40+50=90^0\)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 2

 

xét ΔMNI có

ˆM1+ˆN+ˆI1=1800M1^+N^+I1^=1800 (3 góc trong 1 tam giác)

→ˆM1=180−ˆN−ˆI1=180−60−70→ˆM1=500→M1^=180−N^−I1^=180−60−70→M1^=500

lại có

ˆI1+ˆI2=1800(kề−bù)→ˆI2=180−ˆI1=180−70→ˆI2=1100I1^+I2^=1800(kề−bù)→I2^=180−I1^=180−70→I2^=1100

xét ΔMQI có

ˆM2+ˆQ+ˆI2=1800M2^+Q^+I2^=1800 (3 góc trong 1 tam giác)

→ˆM2=180−ˆQ−ˆI2=180−30−110→ˆM2=400→M2^=180−Q^−I2^=180−30−110→M2^=400

mà 

ˆM1+ˆM2=ˆNMQ→ˆMNQ=40+50=900