Question

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

Answer 1

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