`AB=c=4;AC=b=5;BC=a=6`
`a)` Áp dụng định lý cosin ta có:
`a^2=b^2+c^2-2bc.cosA`
`=>cosA={b^2+c^2-a^2}/{2bc}`
`={5^2+4^2-6^2}/{2.5.4}=1/ 8`
`=>\hat{A}≈82°49'`
`b^2=a^2+c^2-2ac.cosB`
`=>cosB={a^2+c^2-b^2}/{2ac}`
`={6^2+4^2-5^2}/{2.6.4}=9/{16}`
`=>\hat{B}≈55°46'`
`=>\hat{C}=180°-(\hat{A}+\hat{B})≈41°25'`
Vậy `\hat{A}≈82°49'; \hat{B}≈55°46';\hat{C}≈41°25'`
`b)` Áp dụng công thức đường trung tuyến:
`m_a^2={2(b^2+c^2)-a^2}/4`
`={2(5^2+4^2)-6^2}/4={23}/2`
`=>m_a=\sqrt{{23}/2}`$(đvđd)$
`m_b^2={2(a^2+c^2)-b^2}/4`
`={2(6^2+4^2)-5^2}/4={79}/4`
`=>m_b=\sqrt{79}/2`$(đvđd)$
`m_c^2={2(a^2+b^2)-c^2}/4`
`={2(6^2+5^2)-4^2}/4={53}/2`
`=>m_c=\sqrt{{53}/2}`$(đvđd)$
`S_{∆ABC}=1/ 2 ab.sinC`
`≈1/ 2 .6 .5 .sin 41°25'≈9,9`$(đvdt)$
Vậy
`m_a=\sqrt{{23}/2}`$(đvđd)$
`m_b=\sqrt{79}/2`$(đvđd)$
`m_c=\sqrt{{53}/2}`$(đvđd)$
Diện tích $∆ABC$ khoảng $9,9(đvdt)$
