[tex]\it In\ \Delta ABD,\ dreptunghic\ \^in\ A,\ avem \ m(\widehat{ABD})=60^o:2=30^o[/tex]
[tex]\it tg(ABD)=\dfrac{AD}{AB} \Rightarrow tg30^o=\dfrac{AD}{15} \Rightarrow \dfrac{^{5)}\sqrt3}{\ 3}=\dfrac{AD}{15} \Rightarrow \dfrac{5\sqrt3}{15}=\dfrac{AD}{15} \Rightarrow \\ \\ \\ \Rightarrow AD=5\sqrt3\ cm[/tex]
[tex]\it Ducem\ \ CF\perp AB,\ \ F\in\ AB.\\ \\ AFCD-dreptunghi\ \Rightarrow CF=AD=5\sqrt3\ cm\\ \\ In\ \Delta CFB,\ dreptunghic\ \^in\ F,\ avem\ m(\widehat{FBC})=60^o[/tex]
[tex]\it sin(FBC)=\dfrac{CF}{BC} \Rightarrow sin60^o=\dfrac{5\sqrt3}{BC} \Rightarrow \dfrac{\sqrt3}{2}=\dfrac{5\sqrt3}{BC} \Rightarrow BC=\dfrac{2\cdot5\sqrt3}{\sqrt3}=10[/tex]
[tex]\it\ In\ \Delta\ CFB \Rightarrow m(\widehat{BCF})=30^o\ (complementul\ lui\ 60^0) \Rightarrow FB=5\ cm\\ \\ (cu\ T.<30^o)\\ \\ AF=AB-FB=15-5=10\ cm\\ \\ AFCD-dreptunghi \Rightarrow CD=AF=10\ cm[/tex]
[tex]\it \mathcal{P}_{ABCD} =AB+BC+CD+AD=15+10+10+5\sqrt3=35+5\sqrt3\ cm[/tex]
