Question

Cine imi rezolva si mie ex 9.......va rog

Answer 1

Intr-un,triunghi,oarecare,avem,AC=6,AB=4,si,m(∡BAC)=π/3

Teorema,lui,Pitagora,generalizata
BC²=AB²+AC²-2·AB·AC·cosBAC
BC²=16+36-2·4·6·cosπ/3
cosπ/3=1/2

BC²=52-48·1/2
BC²=52-24
BC²=28
BC=2√7

Coboram,CD,inaltimea,din,C,pe,baza,AB
In,ΔCDA,dr.,avem,m(∡CAD)=60°=>m(∡ACD)=30°=>cateta,AD,este,opusa
unghiului,de,30°=>este,jumatate,din,ipotenuza=>AD=3=>DB=1
Cu,t.Pit.avem,CD²=AC²-AD²
CD²=36-9
CD²=27
CD=3√3
In,Δdr.CDB,avem,sinB=CD/CB=3√3/2√7=3√21/14
sinDCB=1/2√7=√7/14