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C

chaugiang81

ta có :
$sinB= \dfrac{AC}{AB} = \dfrac{2}{5}= 0.4 $
$\hat{B}= 23^o 35'$
$cosA= \dfrac{AC}{AB}= \dfrac{2}{5}$
$\hat{A}= 66^o 25'$
$=> \hat{A} - \hat{B}= x-y= 66^o 25' - 23^o 35'= 42^o 50'$
 
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