xet tam giac ABC co \(AB=70cm,AC=60cm\left(gt\right)\)
\(\)ta co: \(BC^2=AB^2+AC^2=>BC=\sqrt{AB^2+AC^2}=\sqrt{70^2+60^2}=10\sqrt{85}\)
theo he thuc luong trong tam giac vuong ABC vuong tai A co:
\(sinB=\frac{AC}{BC}=\frac{66}{10\sqrt{85}}\approx0,72=>\widehat{B}\approx46^0\)
xet tam giac ABC vuong tai A co:
\(\widehat{B}+\widehat{C}=90^0=>\widehat{C}=90^0-\widehat{B}=90^0-46^0=44^0\)
Vay BC = 10\(\sqrt{85}\)
\(\widehat{B}\approx46^0\)
\(\widehat{C}=44^0\)
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