Question

cho tam giác ABC có c=5 ; a= 7 ; góc B=120° Tính: S , AC, ha , R

Answer 1

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinABC\)

\(=\dfrac{1}{2}\cdot5\cdot7\cdot sin120=\dfrac{35\sqrt{3}}{4}\)

Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{5^2+7^2-AC^2}{2\cdot5\cdot7}=cos120=\dfrac{-1}{2}\)

=>\(25+49-AC^2=-35\)

=>\(AC^2=25+49+35=109\)

=>\(AC=\sqrt{109}\)

Kẻ AH\(\perp\)BC

=>\(h_A=AH\)

\(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\)

=>\(\dfrac{1}{2}\cdot AH\cdot7=\dfrac{35\sqrt{3}}{4}\)

=>\(AH\cdot3,5=\dfrac{35\sqrt{3}}{4}\)

=>\(AH=\dfrac{10\sqrt{3}}{4}=\dfrac{5}{2}\sqrt{3}\)

Xét ΔABC có \(\dfrac{AC}{sinB}=2R\)

=>\(2R=\dfrac{\sqrt{109}}{sin120}=\sqrt{109}\cdot\dfrac{2}{\sqrt{3}}\)

=>\(R=\sqrt{\dfrac{109}{3}}=\dfrac{\sqrt{327}}{3}\)