Qua `C` kẻ đường thẳng `d //// AB //// DE`
Ta có:
`AB //// d`
`=> hat{ABC} + hat{C_1} = 180^0`
`=> hat{C_1} = 180^0 - 126^0 = 54^0`
`DE //// d`
`=> hat{CBE} + hat{C_3} = 180^0`
`=> hat{C_3} = 180^0 - 101^0 = 79^0`
Ta có:
`hat{C_1} + hat{C_2} + hat{C_3} = 180^0`
`=> hat{C_2} = 180^0 - 79^0 - 54^0 = 47^0`