Câu 15:
a: Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot cosA\)
=>\(BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot cosA\)
=>Sai
b: \(BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot cosA\)
=>\(BC^2=8^2+5^2-2\cdot8\cdot5\cdot cos60=49\)
=>BC=7(cm)
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot8\cdot5\cdot sin60=10\sqrt{3}\)
=>Đúng
c: Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)
=>\(2R=7:sin60=\dfrac{14\sqrt{3}}{3}\)
=>\(R=\dfrac{7\sqrt{3}}{3}\)
=>Sai
d: Xét ΔBAC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(cosB=\dfrac{8^2+7^2-5^2}{2\cdot8\cdot7}=\dfrac{88}{16\cdot7}=\dfrac{11}{14}\)
Xét ΔBAM có \(cosB=\dfrac{BA^2+BM^2-AM^2}{2\cdot BA\cdot BM}\)
=>\(\dfrac{8^2+4^2-AM^2}{2\cdot8\cdot4}=cosB=\dfrac{11}{14}\)
=>\(80-AM^2=\dfrac{352}{7}\)
=>\(AM^2=80-\dfrac{352}{7}=\dfrac{208}{7}\)
=>\(AM=\sqrt{\dfrac{208}{7}}=\dfrac{4\sqrt{91}}{7}\)
=>Đúng
