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

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

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

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

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

${32^0}$                   

${35^0}$       

$24^\circ$

${90^0}$

$\widehat B + \widehat C = 90^\circ$ 

$\widehat B + \widehat C = 180^\circ$   

$\widehat B + \widehat C = 100^\circ$ 

$\widehat B + \widehat C = 60^\circ$

${40^0}$

${50^0}$

${60^0}$

${100^0}$

${50^0}$       

$80^\circ$    

${100^0}$     

${90^0}$

${40^0}$                                     

${50^0}$

${49^0}$                                      

${98^0}$