Question

Dau funda si mai dau puncte! Urgent va rog

Answer 1

Răspuns:

Răspuns:

$AM=9, AO=\frac{2}{3}*AM=> AO = 6 => OM=3$

$BC=10, BM=MC=\frac{1}{2}*BC=5$

$BO^2=BM^2-OM^2<=>5^2-3^2=25-9=16=>BO=4$

$BO=\frac{2}{3}*BN=>BN=6=>ON=2$

$BA^2=BO^2+AO^2=4^2+6^2=16+36=52=>BA=2\sqrt{13}$

$AN^2=AO^2+ON^2=6^2+2^2=36+4=40=> AN=2\sqrt{10}$

$AN=NC=2\sqrt{10}$

Teorema cosinusului:

$AC^2=AB^2+BC^2-2AB*AC*cos(B)=>cos(B)=\frac{AB^2+BC^2-AC^2}{2AB*BC}$

$=\frac{52+100-160}{2*2\sqrt{13}*10 } =\frac{-8}{40\sqrt{13} }=\frac{-1}{5\sqrt{13} }=\frac{-\sqrt{13} }{65} => cos(ABC)=\frac{-\sqrt{13} }{65}$

Answer 2

$\it Determin\breve am\ \ AB=2\sqrt{13},\ \ AC=4\sqrt{10}\\ \\ AB^2=(2\sqrt{13})^2=4\cdot13=52\\ \\ BC^2=10^2=100\\ \\ AC^2=(4\sqrt{10})^2=16\cdot10=160\\ \\ AB^2+BC^2=52+100=152\approx 160=AC^2$

152 < 160 ⇒ ∡ABC - obtuz, dar foarte aproape de 90°

sin90° = 1 ⇒ sin(ABC) ≈ 0,99

∡ABC - obtuz ⇒ cos(ABC) < 0

Folosim formula fundamentală a trigonometriei:

$\it cos(ABC)=-\sqrt{1-sin^2(ABC)}=-\sqrt{1-0,99^2}=\\ \\ =-\sqrt{1-0,9801}=-\sqrt{0,0199}\approx\ -0,14$