Giải thích các bước giải:
Đặt $a=BC=13, b=AC=15, c=AB=13$
a.Ta có:
$\cos A=\dfrac{AB^2+AC^2-BC^2}{2AB\cdot AC}=\dfrac{15}{26}$
$\to \hat A=\arccos(\dfrac{15}{26})$
$\cos B=\dfrac{BA^2+BC^2-AC^2}{2BA\cdot BC}=\dfrac{113}{338}$
$\to \hat B=\arccos(\dfrac{113}{338})$
$\to \hat C=180^o-\hat A-\hat B= 180^o-\arccos(\dfrac{15}{26})-\arccos(\dfrac{113}{338})$
b.Gọi $m_a, m_b, m_c$ lần lượt là độ dài đường trung tuyến hạ từ $A, B, C$ xuông $BC, AC, AB$
Ta có:
$m_a=\sqrt{\dfrac{2(b^2+c^2)-a^2}{4}}=\dfrac{\sqrt{619}}{2}$
$m_b=\sqrt{\dfrac{2(a^2+c^2)-b^2}{4}}=\dfrac{\sqrt{451}}{2}$
$m_c=\sqrt{\dfrac{2(a^2+b^2)-c^2}{4}}=\dfrac{\sqrt{619}}{2}$
c.Ta có:
$S_{ABC}=\dfrac12bc\sin A$
$\to S_{ABC}=\dfrac12\cdot 15\cdot 13\cdot \sin\arccos(\dfrac{15}{26})$
$\to S_{ABC}=\dfrac{15\sqrt{451}}{4}$
$\to \dfrac12\cdot (a+b+c)\cdot r=\dfrac{15\sqrt{451}}{4}$
$\to \dfrac12\cdot (13+15+13)\cdot r=\dfrac{15\sqrt{451}}{4}$
$\to r=\dfrac{15\sqrt{451}}{82}$
Mặt khác:
$S_{ABC}=\dfrac{abc}{4R}$
$\to \dfrac{15\sqrt{451}}{4}=\dfrac{13\cdot 15\cdot 13}{4R}$
$\to R=\dfrac{169\sqrt{451}}{451}$
d.Ta có:
$S_{ABC}=\dfrac12ah_a=\dfrac12bh_b=\dfrac12ch_c$
$\to ah_a=bh_b=ch_c=2S_{ABC}$
$\to ah_a=bh_b=ch_c= \dfrac{15\sqrt{451}}{2}$
$\to 13h_a=15h_b=13h_c= \dfrac{15\sqrt{451}}{2}$
$\to h_a=h_c=\dfrac{15\sqrt{451}}{26}$
$h_b=\dfrac{\sqrt{451}}{2}$
