Question

[chi nguyen]

cho tam giac ABC co AB=a , BC=a\(\sqrt{7}\)va goc A= 120 do

a) giai tam giac

b)tinh ma, hb, R,r

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Xét ΔBAC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(\dfrac{a^2+AC^2-7a^2}{2\cdot a\cdot AC}=\dfrac{-1}{2}\)

=>\(2\left(AC^2-6a^2\right)=-2a\cdot AC\)

=>\(AC^2-6a^2=AC\cdot-a\)

=>\(AC^2+AC\cdot a-6a^2=0\)

=>AC^2+3*AC*a-2*AC*a-6a^2=0

=>AC(AC+3a)-2a(AC+3a)=0

=>AC=2a

Xét ΔBAC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}=\dfrac{a^2+7a^2-4a^2}{2\cdot a\cdot a\sqrt{7}}=\dfrac{2\sqrt{7}}{7}\)

nên góc B=41 độ

=>góc C=180-120-41=60-41=19 độ

b: \(m_A=\sqrt{\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}}=\sqrt{\dfrac{a^2+4a^2}{2}-\dfrac{7a^2}{4}}=\dfrac{\sqrt{3}}{2}\cdot a\)

\(\dfrac{BC}{sinA}=2\cdot R\)

=>\(2\cdot R=\dfrac{a\sqrt{7}}{sin120}=a\sqrt{7}\cdot\dfrac{2}{\sqrt{3}}\)

=>\(R=a\sqrt{\dfrac{7}{3}}\)