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Trả lời:
$a,$
$\cos A=\dfrac{AB^2+AC^2-BC^2}{2.AB.AC}=\dfrac{12^2+13^2-15^2}{2.12.13}=\dfrac{11}{39}$
$⇒A=73^o37'$
$\cos B=\dfrac{AB^2+BC^2-AC^2}{2.AB.BC}=\dfrac{12^2+15^2-13^2}{2.12.15}=\dfrac{5}{9}$
$⇒B=56^o15'$
$\cos C=\dfrac{AC^2+BC^2-AB^2}{2.AC.BC}=\dfrac{13^2+15^2-12^2}{2.13.15}=\dfrac{25}{39}$
$⇒C=50^o7'$.
$b,$
$m_a=\sqrt{\dfrac{2(AB^2+AC^2)-BC^2}{4}}=\sqrt{\dfrac{2(12^2+13^2)-15^2}{4}}=\dfrac{\sqrt{401}}{2}$
$m_b=\sqrt{\dfrac{2(AB^2+BC^2)-AC^2}{4}}=\sqrt{\dfrac{2(12^2+15^2)-13^2}{4}}=\dfrac{\sqrt{569}}{2}$
$m_c=\sqrt{\dfrac{2(AC^2+BC^2)-AB^2}{4}}=\sqrt{\dfrac{2(13^2+15^2)-12^2}{4}}=\sqrt{161}$.
$c,$
$\cos A=\dfrac{11}{39}⇒\sin A=\dfrac{10\sqrt{14}}{39}$
$⇒S_{ΔABC}=\dfrac{1}{2}.AB.AC.\sin A=\dfrac{1}{2}.12.13.\dfrac{10\sqrt{14}}{39}=20\sqrt{14}$.
