Question

In triunghiul ABC, cu AD _|_ BC, D este in (BC) , avem m(< B ) =30 grade si m ( < C) = 45 grade. Daca AC=18 radical 2, calculati : a)Aria triunghiului ABC, b)lungimile inaltimilor triunghiului dat.
Va rog!!!Urgent!!!

Answer 1

a)Am,desenat,ΔABC,cu,baza,BC,pe,orizontala.
Am,coborat,AD_l_BC(D∈BC)
In,Δdr.ADB,avem,m(∡ABD)=30°(ipoteza)=>cateta,opusa,unghiului,de,30°,este,
jumatate,din,ipotenuza=>AD=AB/2=>AB=2AD
In,ΔdrADC,avem,m(∡ACD)=45°=>m(∡CAD)=45°=>ΔdrADC-isoscel
=>AD=DC=x
Cu,t.Pit.,avem,x²+x²=648=>2x²=648=>x²=324=>AD=DC=18
=>AB=2x18=>AB=36
Cu,t.Pit.,aflam,BD
AB²=BD²+AD²
1296=BD²+324
BD²=972
BD=18√3=>BC=18√3+18=18(√3+1)

Aria=18x18(√3+1)/2=162(√3+1)

b)CE_l_AB,(E∈AB)
In,Δdr.BEC,cateta,CE,este,opusa,unghiului,de,30°=>este,jumatate,din,ipotenuza,BC=>CE=9

BF_l_AC(F∈AC)
In,Δdr.BFC,avem,m(∡FCB)=45°(ipoteza)=>m(∡FBC)=45°=>Δdr.BFC,isoscel=>BF=FC=a
Cu,t.Pit.,avem,a²+a²=(18√3+18)²
2a²=972+324√3+324
a²=1296+324√3
a²=324(4+√3)
BF=18√(4+√3)