a, Vì a⊥AB và b⊥AB nên a//b
b, \(\widehat{C_2}=180^0-\widehat{C_1}=133^0\left(kề.bù\right)\)
Vì a//b nên \(\left\{{}\begin{matrix}\widehat{C_2}=\widehat{D_3}=133^0\left(slt\right)\\\widehat{C_1}=\widehat{D_2}=47^0\left(đv\right)\end{matrix}\right.\)
Ta có \(\widehat{D_3}=\widehat{D_1}=133^0\left(đối.đỉnh\right)\)






