Question

Repede ofer 28 de puncte si coroana.Punctele A,B,C sunt coliniare in aceasta ordine,AB=2cm,BC=8cm si DB perpendicular AC,DB=4cm.
a)Calculati sin unghiul ADB + sin unghiul BDC.
b)Aflati perimetrul triunghiului ADC.​

Answer 1

In ΔDBC, m(∡DBA) = 90° => AD²=AB²+BD²

=> AD²= 20

=>AD= 2√5

sin(∡ADB) = $\frac{AB}{AD}$

=>sin(∡ADB) = $\frac{2}{2\sqrt{5} }$

=>sin(∡ADB) = $\frac{\sqrt{5} }{5}$

In Δ BDC, m (∡DBC) = 90° => DC²= DB²+BC²

=>DC²=80

=>DC=4$\sqrt{5}$

sin(∡BDC) = $\frac{BC}{DC}$

=>sin(∡BDC)=$\frac{8}{4\sqrt{5} }$

=>sin(∡BDC) = $\frac{2\sqrt{5} }{5}$

sin(∡ADB)+sin(∡BDC)= $\frac{3\sqrt{5} }{5}$

b) P Δ ADC = 2√5+4√5+10 = 6√5 + 10

=2(3√5+5) cm