Question

Va rog repede am nevoie de rezolvare dau coroana

Answer 1

$cos(ACB)=\dfrac{12}{24}=\dfrac{1}{2} \Rightarrow \widehat{ACB}=60^o\\ \\ \\ CM - bisectoare \Rightarrow \widehat{ACB}=60^o:2=30^o\\ \\ cos(ACM)=cos30^o=\dfrac{12}{AM} \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{12}{CM} \Rightarrow CM=\dfrac{^{\sqrt3)}2\cdot12}{\sqrt3}=\dfrac{2\cdot12\sqrt3}{3}=8\sqrt3$