Question

Fie ABC triunghi m(<A)=120° AB=a, BC=a+2, CA=a+1 unde a apartine (0,+infinit).Determunati valoarea lui a.(p.s: Raspunsul trebuie sa dea 3/2)

Answer 1

Teoria cosinusului aplicata pentru unghiul A este
$\cos{A}=\frac{AB^{2}+AC^{2}-BC^{2}}{2ABAC}=\frac{a^{2}+(a+1)^{2}-(a+2)^{2}}{2a(a+1)}=\frac{a^{2}+a^{2}+2a+1-a^{2}-4a-4}{a(a+1)}=\frac{a^{2}-2a-3}{a(a+1)}=\frac{a^{2}+a-3a-3}{a(a+1)}=\frac{a(a+1)-3(a+1)}{a(a+1)}=\frac{(a-3)(a+1)}{a(a+1)}=\frac{a-3}{a}$
Stim ca in general
$\cos{(\pi-x)}=-\cos{x}$ atunci putem scrie
$\cos{120}=\cos{(\pi-60)}=-\cos{60}=-\frac{1}{2}$
Atunci avem
$\cos{A}=\frac{a-3}{a}=-\frac{1}{2}\Rightarrow 2a-6=-a\Rightarrow 3a=6\Rightarrow a=2$