Question

In triunghiul ABC se considera inaltimea AD⊥BC,D∈BC.Stind ca AD=12 cm ,cosC=frac{5}{13} si sinB=\frac{3}{5} ,calculati lungimea laturii [BC]

Answer 1

Am desenat triunghiul ABC cu unghiul A in sus si baza BC pe orizontala.
Avem ΔABD dr, creat de inaltimea AD_l_ BC
sin B= cateta opusa/ipotenuza=12/AB=3/5=>AB=20
Segm. BD il aflam cu t.Pit. AB²=AD²+BD²=>BD²=AB²-AD²=>BD²=400-144
BD²=256=>BD=16 cm
cos C=cateta alaturata/ipotenuza=5/13=CD/AC=>CD=5AC/13
AD²+CD²=AC²
144+25AC²/169=AC²
24336+25AC²=169AC²
24336=144AC²=>AC²=169 =>AC=13 cm
DC²=AC²-AD²
DC²=169-144
DC=5 cm 

=>BC=21 cm

Answer 2

sin(B) = 12/AB = 3/5

AB=20

pitagora in ABD

BD^2 = AB^2 - AD^2

BD^2 = 400 - 144

BD=16

cos(C) = DC/AC = 5/13, si in plus cu pitagora in ADC avem:

DC^2 = AC^2 - AD^2 aici inlocuim pe AC=13 x DC/5

25 x DC^2 = 13^2 x DC^2 - 12^2 x 25

DC=5

BC = BD+DC = 16+5=21