In ΔDBC, m(∡DBA) = 90° => AD²=AB²+BD²
=> AD²= 20
=>AD= 2√5
sin(∡ADB) = [tex]\frac{AB}{AD}[/tex]
=>sin(∡ADB) = [tex]\frac{2}{2\sqrt{5} }[/tex]
=>sin(∡ADB) = [tex]\frac{\sqrt{5} }{5}[/tex]
In Δ BDC, m (∡DBC) = 90° => DC²= DB²+BC²
=>DC²=80
=>DC=4[tex]\sqrt{5}[/tex]
sin(∡BDC) = [tex]\frac{BC}{DC}[/tex]
=>sin(∡BDC)=[tex]\frac{8}{4\sqrt{5} }[/tex]
=>sin(∡BDC) = [tex]\frac{2\sqrt{5} }{5}[/tex]
sin(∡ADB)+sin(∡BDC)= [tex]\frac{3\sqrt{5} }{5}[/tex]
b) P Δ ADC = 2√5+4√5+10 = 6√5 + 10
=2(3√5+5) cm
