Question

Answer 1

Xét ΔACB có

\(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

=>\(15^2+21^2-AB^2=2\cdot15\cdot21\cdot cos80\)

=>\(666-AB^2=630\cdot cos80\)

=>\(AB^2=666-630\cdot cos80\)

=>\(AB=\sqrt{666-630\cdot cos80}\simeq23,59\left(m\right)\)