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2014-04-24T00:26:08+03:00
ΔADB mas<A=90
sin<B=30=  1/2
sin<B=AD/DB     1/2=6/DB         ⇒DB=6*2=12
DB=AC=12
AB²=AD²+DB²=  6²+  12²=  36+144=180   ⇒AB=6√5
AB=DC=6√5
ΔADC, mas<D=90  fie DN_/_AC⇒  d(D,AC)=DN
DN-h   ⇒DN=AD*DC/AC=   6*6√5/12=  6√5/2=  3√5