Câu 3:
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{14^2+15^2-13^2}{2\cdot14\cdot15}=\dfrac{3}{5}\)
=>\(sinBAC=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)
Xét ΔABC có \(\dfrac{BC}{sinBAC}=2R\)
=>\(2R=13:\dfrac{4}{5}=13\cdot\dfrac{5}{4}=\dfrac{65}{4}\)
=>\(R=\dfrac{65}{8}\)
=>Chọn A
Câu 4:
\(S_{BAC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinBAC=\dfrac{1}{2}\cdot4\cdot5\cdot sin150=5\)
=>Chọn B
Câu 3
Theo Hê rông
\(S_{ABC}=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}=84\)
\(S_{ABC}=\dfrac{AB.BC.AC}{4R}\Rightarrow R=\dfrac{AB.BC.AC}{4S_{ABC}}=\dfrac{65}{8}\)
=> A
Câu 4
\(S_{ABC}=\dfrac{1}{2}.BC.AB.sin150^0=5\)
=> B
