`B1:`
$\text{Tứ giác ABCD có:}$
$\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o$
$⇒\widehat{A}=\widehat{C}=\widehat{D}=\dfrac{360^o-\widehat{B}}{3}=\dfrac{360^o-96^o}{3}=88^o$
`B2:` $\text{(hình dưới)}$
$\text{Ta có:}$
$\widehat{B_1}+\widehat{B_2}=180^o$
$⇒\widehat{B_1}=180^o - \widehat{B_2} =180^o - 90^o=90^o$
$\widehat{C_1}+\widehat{C_2}=180^o$
$⇒\widehat{C_1}=180^o - \widehat{C_2} =180^o - 110^o=70^o$
$\text{Tứ giác ABCD có:}$
$\widehat{A}+\widehat{B_1}+\widehat{C_1}+\widehat{D}=360^o$
$⇒\widehat{D}= 360^o - \widehat{A}-\widehat{B_1}-\widehat{C_1}$
`=360^o - 50^o - 90^o - 70^o =150^o`
$⇒\widehat{D}=150^o$