Bài giải:

Gọi tứ giác là: ABCD.

Các góc ngoài tại đỉnh A, B, C, D lần lượt là: A₁ ; B₁ ; C₁; D₁

Ta có: $\widehat{A}+\widehat{{{A}_{1}}}={{180}^{0}}$

$\widehat{B}+\widehat{{{B}_{1}}}={{180}^{0}}$

$\widehat{C}+\widehat{{{C}_{1}}}={{180}^{0}}$

$\widehat{D}+\widehat{{{D}_{1}}}={{180}^{0}}$

Suy ra: $\widehat{A}+\widehat{{{A}_{1}}}+\widehat{B}+\widehat{{{B}_{1}}}+\widehat{C}+\widehat{{{C}_{1}}}+\widehat{D}+\widehat{{{D}_{1}}}={{180}^{0}}+{{180}^{0}}+{{180}^{0}}+{{180}^{0}}$

$\Rightarrow $ $\widehat{A}+\widehat{{{A}_{1}}}+\widehat{B}+\widehat{{{B}_{1}}}+\widehat{C}+\widehat{{{C}_{1}}}+\widehat{D}+\widehat{{{D}_{1}}}={{720}^{0}}$

$\Rightarrow $$\widehat{{{A}_{1}}}+\widehat{{{B}_{1}}}+\widehat{{{C}_{1}}}+\widehat{{{D}_{1}}}=720-(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D})$ 

$\Rightarrow $$\widehat{{{A}_{1}}}+\widehat{{{B}_{1}}}+\widehat{{{C}_{1}}}+\widehat{{{D}_{1}}}={{720}^{0}}-{{360}^{0}}={{360}^{0}}$           (vì $\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}={{360}^{0}}$ )

Vậy đáp án đúng là: A