Question

Giúp mk câu1 vs ạ. Mình cảm ơn

Answer 1

1:

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

=>\(\dfrac{25^2+20^2-BC^2}{2\cdot25\cdot20}=\dfrac{1}{2}\)

=>\(1025-BC^2=500\)

=>BC^2=525

=>\(BC=5\sqrt{21}\left(cm\right)\)

\(S_{ABC}=\dfrac{1}{2}\cdot20\cdot25\cdot sin60\simeq216.506\left(cm^2\right)\)