Question

Se considera triunghiul isoscel ABC cu BC = 10 cm , D = prBC A si masura ungiului BAD = 60 grade. Aflati lungimea inaltimii [ AD ] si perimetrul triunghiului.

Answer 1

[tex]In\ \triangle ABD\rightarrow tg(\ \textless \ BAD)=\frac{BD}{AD}\\ \sqrt{3}=\frac{5}{AD}\\ AD=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\\ m(\ \textless \ ABD)=30^0\rightarrow(T\ \textless \ 30^0) AB=2AD=\frac{10\sqrt{3}}{3}=AC\\ P_{\triangle ABC}=AB+AC+BC=\frac{20\sqrt{3}}{3}+10 [/tex]

Answer 2

ABC tr isoscel, AD inaltime AD bisectoare, AD mediatoare, BD=10/2=5cm
m∡BAD=60°, m∡ABD=30°
vezi atas pt AD=(5√3)/3 cm  si perimetru=((20√3)/3 +10)cm