Đáp án:

a)  $BC=7$ , $\widehat{B}\approx 81{}^\circ 47'$ , $\widehat{C}\approx 38{}^\circ 13'$

b)

${{S}_{\Delta ABC}}=10\sqrt{3}$

$R=\dfrac{7\sqrt{3}}{3}$

$r=\sqrt{3}$

$AH=\dfrac{20\sqrt{3}}{7}$

$AM=\dfrac{\sqrt{129}}{3}$

Giải thích các bước giải:

$AB=5\,,\,AC=8\,,\,\widehat{A}=60{}^\circ $

a)

$\bullet \,\,\,\,\,B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB.AC.\cos \widehat{A}$

$\to B{{C}^{2}}={{5}^{2}}+{{8}^{2}}-2.5.8.\cos 60{}^\circ $

$\to B{{C}^{2}}=49$

$\to BC=7$

$\bullet \,\,\,\,\,\cos B=\dfrac{A{{B}^{2}}+B{{C}^{2}}-A{{C}^{2}}}{2AB.BC}$

$\to \cos B=\dfrac{{{5}^{2}}+{{7}^{2}}-{{8}^{2}}}{2.5.7}$

$\to \cos B=\dfrac{1}{7}$

$\to \widehat{B}\approx 81{}^\circ 47'$

$\bullet \,\,\,\,\,\cos C=\dfrac{B{{C}^{2}}+A{{C}^{2}}-A{{B}^{2}}}{2BC.AC}$

$\to \cos C=\dfrac{{{7}^{2}}+{{8}^{2}}-{{5}^{2}}}{2.7.8}$

$\to \cos C=\dfrac{11}{14}$

$\to \widehat{C}\approx 38{}^\circ 13'$

b)

Nữa chu vi $p=\dfrac{AB+AC+BC}{2}=\dfrac{5+8+7}{2}=10$

Diện tích:

${{S}_{\Delta ABC}}=\sqrt{p\left( p-AB \right)\left( p-AC \right)\left( p-BC \right)}$

$\to {{S}_{\Delta ABC}}=\sqrt{10\left( 10-5 \right)\left( 10-8 \right)\left( 10-7 \right)}$

$\to {{S}_{\Delta ABC}}=10\sqrt{3}$

Bán kính đường tròn ngoại tiếp

${{S}_{\Delta ABC}}=\dfrac{AB.AC.BC}{4R}$

$\to R=\dfrac{AB.AC.BC}{4{{S}_{\Delta ABC}}}$

$\to R=\dfrac{5.8.7}{4.10\sqrt{3}}$

$\to R=\dfrac{7\sqrt{3}}{3}$

Bán kính đường tròn nội tiếp

$S=p.r$

$\to r=\dfrac{S}{p}$

$\to r=\dfrac{10\sqrt{3}}{10}$

$\to r=\sqrt{3}$

Đường cao $AH$

${{S}_{\Delta ABC}}=\dfrac{1}{2}AH\cdot BC$

$\to AH=\dfrac{2{{S}_{\Delta ABC}}}{BC}$

$\to AH=\dfrac{2.10\sqrt{3}}{7}$

$\to AH=\dfrac{20\sqrt{3}}{7}$

Trung tuyến $AM$

$A{{M}^{2}}=\dfrac{A{{B}^{2}}+A{{C}^{2}}}{2}-\dfrac{B{{C}^{2}}}{4}$

$\to A{{M}^{2}}=\dfrac{{{5}^{2}}+{{8}^{2}}}{2}-\dfrac{{{7}^{2}}}{4}$

$\to A{{M}^{2}}=\dfrac{129}{4}$

$\to AM=\dfrac{\sqrt{129}}{3}$