a) Aplicam Teorema lui Pitagora in triunghiul dreptunghic ABD si obtinem
[tex]BD= \sqrt{AB^2+AD^2}= \sqrt{3^2+4^2}= \sqrt{9+16}= \sqrt{25}=5\ cm [/tex]
b) In triunghiul dreptunghic ABD avem
[tex]\sin\angle (ABD)= \frac{AD}{BD}= \frac{4}{5}[/tex]
c)
In triunghiul dreptunghic ABM avem
[tex]\sin\angle (ABM)= \frac{4}{5} \\ Dar\ \sin\angle (ABM) = \frac{AM}{AB} \\ Deci\ \frac{AM}{AB}= \frac{4}{5}\Rightarrow AM= \frac{4\cdot AB}{5}= \frac{4\cdot3}{5}= \frac{12}{5}\ cm.[/tex]
Stim ca
[tex]\sin^2\angle B+\cos^2\angle B=1 \\ Deci\ \cos\angle B= \sqrt{1-\sin^2\angle B}= \sqrt{1- \frac{16}{25} }= \frac{3}{5} \\ \cos\angle B= \frac{MB}{AB}= \frac{MB}{3}= \frac{3}{5} \\ MB= \frac{9}{5}\ cm \\ Rotunjit\ la\ cel\ mai\ apropiat\ nr\ intreg\ este\ 2. [/tex]
