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2014-04-19T19:19:31+03:00
Mas<BAD=mas<BAC-mas<CAD=90-60=30
ΔBAD, mas<D=90
mas<BAD=30                ⇒BD=AB/2=100/2=50
⇒mas<ABD=90-30=60

ΔABC,mas<A=90    AB²=BD*BC             100²=50*BC   10000=50*BC   BC=10.000/50=200
1 5 1
2014-04-19T19:20:31+03:00
∧BAD=90-∧DAC=90-60=30 si ∧ADB=90⇒∧ABD=180-90-30=60
intr-un tr dreptunghic cateta care se opune unghiului de 30 de grade = 1/2 din ipotenuza
BD=AB/2=100/2=50 m
Aplici th catetei
AB²=BD.BC
100²=50.BC
b
BC=10000/50=200 m