Question

giải tam giác ABC biết Â = 90 độ, AB = 5cm, AC = 12cm

 

Answer 1

BC=13cm

\(\sin\widehat{B}=\dfrac{12}{13}\)

\(\Leftrightarrow\widehat{B}=67^0\)

hay \(\widehat{C}=23^0\)

Answer 2

\(BC=\sqrt{AB^2+AC^2}=13\left(cm\right)\left(pytago\right)\\ \sin B=\dfrac{AC}{BC}=\dfrac{12}{13}\approx67^0\\ \Rightarrow\widehat{B}\approx67^0\\ \widehat{C}=90^0-\widehat{B}\approx23^0\)