Bài giải:

Xét tứ giác ABCD ta có:

$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}={{360}^{0}}$ (tổng các góc trong tứ giác)

$\Rightarrow \widehat{A}+\widehat{B}={{360}^{0}}-\widehat{C}-\widehat{D}$

Mà: $\widehat{C}={{100}^{0}},\widehat{D}={{40}^{0}}$ (giả thiết)

$\Rightarrow \widehat{A}+\widehat{B}={{360}^{0}}-{{100}^{0}}-{{40}^{0}}={{220}^{0}}$

$\Rightarrow \widehat{A}={{220}^{0}}-\widehat{B}$

Thay $\widehat{A}={{220}^{0}}-\widehat{B}$ vào $\widehat{A}-\widehat{B}={{10}^{0}}$ ta có:

${{220}^{0}}-\widehat{B}-\widehat{B}={{10}^{0}}$

$\Rightarrow $ ${{220}^{0}}-2\widehat{B}={{10}^{0}}$

$\Rightarrow $$\widehat{B}={{105}^{0}}$ 

Suy ra: $\widehat{A}={{105}^{0}}+{{10}^{0}}={{115}^{0}}$

Vậy: $\widehat{A}={{115}^{0}}$