\(BC=\sqrt{AB^2+AC^2}=\sqrt{3,5^2+4,2^2}=\dfrac{7\sqrt{61}}{10}cm\left(Pyatgo\right)\)
\(\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{3,5}{\dfrac{7\sqrt{61}}{10}}\simeq39^0\)
\(\Rightarrow\widehat{B}=\widehat{A}-\widehat{C}=90^0-39^0=51^0\)
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