Vì `AB //// CD`
`=>` \(\left\{ \begin{array}{l}\hat{A} + \hat{D} = 180^0\\\hat{B} + \hat{C} = 180^0\end{array} \right.\)
Ta có:
`hat{A} = 4hat{D}`
Mà: `hat{A} + hat{D} = 180^0`
`=> 4hat{D} + hat{D} = 180^0`
`=> hat{D} = 36^0`
`=> hat{A} = 4.36^0 = 144^0`
Ta có:
`hat{C} - hat{B} = 40^0`
`=> hat{C} = hat{B} + 40^0`
Mặt khác: `hat{B} + hat{C} = 180^0`
`=> hat{B} + hat{B} + 40^0 = 180^0`
`=> 2hat{B} = 140^0`
`=> hat{B} = 70^0`
`=> hat{C} = 70^0 + 40^0 = 110^0`
`=> hat{C} + hat{D} = 110^0 + 36^0 = 146^0`
