Question

va rog. Pls pls pls pls pls pls​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it a)\ tg{\widehat{ABD}}=\dfrac{cateta\ \ opus\breve a}{cateta\ \ al\breve a turat\breve a}=\dfrac{AD}{BD}=\dfrac{4}{2}=2$

$\it b)\ \Delta ABC-dr,\ \hat D=90^0 \stackrel{T.P.}{\Longrightarrow}\ AB^2=AD^2+BD^2=4^2+2^2=16+4=20\\ \\ AB=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5\ cm$

c)

$\it Cu\ teorema\ catetei,\ avem:\\ \\ AB^2=BD\cdot BC \Rightarrow BC=\dfrac{AB^2}{BD}=\dfrac{20}{2}=10\ cm$