Răspuns :
[tex]\displaystyle \\ \text{Se da: } \\ \Delta ABC ~~cu ~~\ \textless \ A=90^o \\ AB = 18 ~cm \\ \ \textless \ B = 30^o \\ \\ \text{Se cere: } \\ Perimetrul ~(P) = ~? \\ \\ \text{Rezolvare: } \\ \cos B = \frac{AB}{BC} \\ \\ BC = \frac{AB}{\cos B} =\frac{18}{\cos 30^o} =\frac{18}{ \frac{ \sqrt{3} }{2} } =\frac{18\times 2}{\sqrt{3} } =\frac{18\times 2\times \sqrt{3} }{3} =\boxed{12\sqrt{3}~cm} \\ \\ AC = \sqrt{BC^2 - AB^2} = \sqrt{(12\sqrt{3})^2 - 18^2} = [/tex]
[tex]=\sqrt{12^2 \times 3 - 18^2} = \sqrt{144 \times 3 - 324} = \\ \\ =\sqrt{432 - 324} = \sqrt{108} = \sqrt{36 \times 3} = \boxed{6\sqrt{3}~cm} \\ \\ P = AB + BC + AC = \\ \\ =18 + 12\sqrt{3} +6\sqrt{3} = 18+18\sqrt{3} = \boxed{18(1+\sqrt{3})~cm} [/tex]
