a: Ta có: MN⊥MN
MH⊥HK
Do đó: MN//HK
b: Ta có: \(\hat{N_1}+\hat{MNx}=180^0\) (hai góc kề bù)
=>\(\hat{N_1}=180^0-112^0=68^0\)
Ta có: \(\hat{N_1}+\hat{N_2}=180^0\) (hai góc kề bù)
=>\(\hat{N_2}=180^0-68^0=112^0\)
Ta có: MN//HK
=>\(\hat{K_2}=\hat{N_1}\) (hai góc so le trong)
=>\(\hat{K_2}=68^0\)
Ta có: \(\hat{K_2}+\hat{K_3}=180^0\) (hai góc kề bù)
=>\(\hat{K_3}=180^0-68^0=112^0\)
