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

\(\Delta ABC = \Delta KOH\) 

\(\Delta ABC = \Delta HOK\)   

\(\Delta ABC = \Delta OHK\) 

\(\Delta ABC = \Delta OKH\)

\(\Delta ABC = \Delta DEF\) 

\(\Delta ABC = \Delta EFD\)   

\(\Delta ABC = \Delta FDE\) 

\(\Delta ABC = \Delta DFE\)

\({40^0}\)                                     

\({50^0}\)

\({49^0}\)                                      

\({98^0}\)

\(\Delta ABC = \Delta CDA\)

\(\widehat {ABC} = \widehat {CDA}\)        

\(\widehat {BAC} = \widehat {DAC}\)        

\(\widehat {BCA} = \widehat {DAC}\)

\(\widehat B = {60^0}\)                                     

\(\widehat B = {90^0}\)

\(\widehat B = {40^0}\)                                   

\(\widehat B = {80^0}\)

\({40^0}\)                                     

\({50^0}\)       

\({49^0}\)                                      

 \({60^0}\)