Cho tam giác $ABC$  có \(\widehat A = {96^0},\widehat C = {50^0}\). Số đo góc $B$ là:

\(\widehat {BAD} + \widehat {ABD} + \widehat {ADB} = 180^\circ \)

\(\widehat {CAD} + \widehat {BAD} + \widehat {BAC} = 180^\circ \)

\(\widehat {CAD} + \widehat {ADC} + \widehat {ACB} = 180^\circ \)

\(\widehat {BAC} + \widehat {ACD} + \widehat {ABD} = 180^\circ \)

\({32^0}\)                   

\({35^0}\)       

\(24^\circ \)

\({90^0}\)

\(\widehat B + \widehat C = 90^\circ \) 

\(\widehat B + \widehat C = 180^\circ \)   

\(\widehat B + \widehat C = 100^\circ \) 

\(\widehat B + \widehat C = 60^\circ \)

\({40^0}\)

\({50^0}\)

\({60^0}\)

\({100^0}\)

\({50^0}\)       

\(80^\circ \)    

\({100^0}\)     

\({90^0}\)

\({40^0}\)                                     

\({50^0}\)

\({49^0}\)                                      

\({98^0}\)