Ta có : ∠A+∠B+∠C+∠D= $360^{0}$
hay : $120^{0}$ + $100^{0}$ +∠C + ∠D= $360^{0}$
⇔$220^{0}$ +∠C + ∠D= $360^{0}$
⇔∠C + ∠D=$360^{0}$ - $220^{0}$
⇔∠C + ∠D= $140^{0}$
Áp dụng t/c tổng - hiệu:
∠C + ∠D= $140^{0}$
∠C - ∠D= $20^{0}$
⇒ ∠C = (140+20) : 2 = $80^{0}$
⇒ ∠D =(140-20) : 2 = $60^{0}$
Vậy góc C =$80^{0}$
góc D =$60^{0}$