Question

Fie ∆ABC dreptunghic si [AD] înaltimea corespunzatoare ipotenuzei [BC] D apartine [BC].Se cere :a) P ∆ABC,daca sinus(BAD)=3pe5 si BD=9cm

Answer 1

In triunghiul ADB:

sin(AOB) = 3/5 => sin(AOB) = cat.op/ip = 9/AB = 3/5 => AB=15

sin^2 + cos^2 =1

sin(AOB) = 3/5 => cos(AOB)^2=1-sin(AOB)^2=1-(3/5)^2 => cos(AOB) = 4/5

cos(AOB) = AD/AB = AD/15, dar cos(AOB) = 4/5 => AD/15=4/5 => AD=12  

Teorema inaltimii :AD=(cat1*cat2)/ip => 12=15*AC/BC => BC=(15/12)/AC   (1)

Teorema catetei : AC^2 = CD*BC = (BC-9)*BC; inlocuim BC din (1)

AC^2=(5/4*AC -9)*5/4*AC => 16*AC^2=25*AC^2-4*45 => 9*AC^2 = 4*5*9*AC =>

AC=20, iar din(1) : BC=5/4*20 => BC = 25

Perim(ABC) = 15+20+25 = 60