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Calculati perimetrul triunghiului ABC cu măsura unghiului B=90°,BC=3√5 cm și sinA=√5 pe 3.

Răspuns :

[tex]BC=3\sqrt{5}[/tex]
[tex]sinA=\frac{BC}{AC}[/tex]
[tex]\frac{BC}{AC}=\frac{\sqrt{5}}{3}[/tex]
[tex]3BC=AC\sqrt{5}[/tex]
[tex]3BC=\sqrt{AB^{2}+BC^{2}}\cdot\sqrt{5}[/tex]
[tex]3BC=\sqrt{5(AB^{2}+BC^{2})}[/tex]
[tex](3BC)^{2}=5AB^{2}+5BC^{2}[/tex]
[tex]9BC^{2}=5AB^{2}+5BC^{2}[/tex]
[tex]4BC^{2}=5AB^{2}[/tex]
[tex]4\cdot3\sqrt{5}=5AB^{2}[/tex]
[tex]12\sqrt{5}=5AB^{2}[/tex]
[tex]12=AB^{2}\sqrt{5}[/tex]
[tex]AB^{2}=\frac{12}{\sqrt{5}}[/tex]
[tex]AB=\sqrt{\frac{12\sqrt{5}}{5}}[/tex]
[tex]AB=\frac{\sqrt{12\sqrt{5}}}{\sqrt{5}}[/tex]
[tex]AB=\frac{\sqrt{12\sqrt{5}}\cdot\sqrt{5}}{5}[/tex]
[tex]AB=\frac{\sqrt{12\sqrt{5}\cdot5}}{5}[/tex]
[tex]AB=\frac{\sqrt{60\sqrt{5}}}{5}[/tex]
[tex]AC^{2}=AB^{2}+BC^{2}[/tex]
[tex]AC^{2}=\Big(\sqrt{\frac{12\sqrt{5}}{5}}\Big)\Up^{2}+(3\sqrt{5})^{2}[/tex]
[tex]AC^{2}=\frac{12\sqrt{5}}{5}+9\cdot5[/tex]
[tex]AC^{2}=\frac{12\sqrt{5}}{5}+45[/tex]
[tex]AC^{2}=\frac{12\sqrt{5}}{5}+\frac{225}{5}[/tex]
[tex]AC^{2}=\frac{12\sqrt{5}+225}{5}[/tex]
[tex]AC=\sqrt{\frac{12\sqrt{5}+225}{5}}[/tex]
[tex]AC=\frac{\sqrt{12\sqrt{5}+225}}{\sqrt{5}}}[/tex]
[tex]AC=\frac{\sqrt{12\sqrt{5}+225}\cdot\sqrt{5}}{5}[/tex]
[tex]AC=\frac{\sqrt{(12\sqrt{5}+225)\cdot5}}{5}[/tex]
[tex]AC=\frac{\sqrt{60\sqrt{5}+1125}}{5}[/tex]
[tex]P_{ABC}=AB+BC+AC=[/tex]
[tex]=\frac{\sqrt{60\sqrt{5}}}{5}+3\sqrt{5}+\frac{\sqrt{60\sqrt{5}+1125}}{5}=[/tex]
[tex]=\frac{\sqrt{60\sqrt{5}}}{5}+\frac{15\sqrt{5}}{5}+\frac{\sqrt{60\sqrt{5}+1125}}{5}=[/tex]
[tex]=\frac{\sqrt{60\sqrt{5}}+15\sqrt{5}+\sqrt{60\sqrt{5}+1125}}{5}[/tex]