Question

Ajutor exerctiile 1,2 și 3, vă rog!

Answer 1

1.

OM⊥AB

a)(∡AOM)=30°

AM=1/2=6 (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

2.

R = 18 cm

a)m(∡AOB)=75°

masura arcului mic AB=75°

b)m(∡OBA)=50 °

ΔAOB isoscel deoarece AO=OA=raza ⇒m(∡OBA)=m(∡OAB)=50 °

⇒ m(∡AOB)=180°- 50 - 50=80°

deci masura arcului mic AB=80°

c)AB = 18√2 cm triunghiul AOB isoscel

OD⊥AB

AD=AB/2=18√2/2=9√2

sin ∡AOD=AD/AO=9√2/18=√2/2

m(∡AOD)=45°, m(∡AOB) = 45°×2 = 90°

3.

mas. arc AB = m(∡ AOB) = (40/100 )× 360° = 144°