Hình minh họa:
Giải:
Có: \(S_{\Delta ABC}=\dfrac{1}{2}\cdot AB\cdot BC\cdot\sin\left(\widehat{B}\right)=33,82911734\)
Kẻ AH _|_ BC
Có: \(\widehat{B_1}=180^o-\widehat{ABC}=180^o-120^o=60^o\)
Ta có: \(\sin\widehat{B_1}=\dfrac{AH}{AB}\)
\(\Rightarrow AH=AB\cdot\sin\widehat{B}=\dfrac{25\sqrt{3}}{8}\)
\(\Rightarrow BH=\sqrt{AB^2-AH^2}=3,125\)
\(\Rightarrow HC=12,5+3,125=15,625\)
\(\Rightarrow AC=\sqrt{AH^2+HC^2}=\dfrac{25\sqrt{7}}{4}\)
Kẻ DH1 _|_ BC
Tam giác ABC có BD là p/g góc ABC nên:
\(\dfrac{AD}{AB}=\dfrac{CD}{BC}=\dfrac{AD+CD}{AB+BC}=\dfrac{\dfrac{25\sqrt{7}}{4}}{6,25+12,5}=0,8819171037\)
\(\Rightarrow AD=\dfrac{25\sqrt{7}}{12};DC=11,02396379\)
Có: \(DH_1=sin\widehat{C}\cdot DC=\dfrac{AH}{AC}\cdot DC=3,608439218\)
\(\Rightarrow BD=\sin\widehat{BDH_1}\cdot DH_1=\sin\left(60^o\right)\cdot DH_1=4,166666708\)
Vậy...
