OM⊥AB
a)
m(∡AOM)=30°
AM=1/2=6 ( aici am teorema de 30 ° )
AM=12 cm
b)
m(∡AOM)=60°
sin60°=AM/OA⇒
⇒√3/2=12/OA ⇒
⇒OA=12*√3/2 ⇒
⇒OA=6√3cm
AB=12√3cm
c)
Teorema lui Pitagora in ΔAOM dr in O ⇔
⇔ i²=c1²+c2² ⇔
⇔AM²=12²-8² ⇔
⇔AM²=144-64=80 ⇔
⇔AM=√80 ⇔
⇔AM=4√5 cm
AB=8√5 cm
sin(∡AOM)=√3/2
d)
m(∡AOM)=60°
sin(∡AOM)=AM/12=√3/2⇔
⇔ AM=12√3/2⇔
⇔AM=6√3cm
AB=12√3cm
e)
m(∡AOB)=90°
AB=AO√2 = 12√2 cm
Sper ca te am ajutat!!!!
