Question

Ajutor, dau coroana!!!!!!!

Answer 1

1) a) m(∡AOB)+m(∡AOC)=180° ⇔ 90°+3x=180° ⇒ x=30°.

m(∡AOB)=m(∡COD)=120°.

m(∡AOC)=m(∡BOD)=60°.

1) b) m(∡AOM)+m(∡DOM)=m(∡AOD) ⇒ m(∡DOM)=m(∡AOD)-m(∡AOM)=

120°-90°=30°.

(ON=bisect.∡BOC ⇔ m(∡BON)=m(∡CON)=1/2∡AOC=60°.

Asadar, m(∡MON)=m(∡BON)+m(∡DOM)+m(∡BOD)=60°+30°+60°=150°.

2) a) m(∡AOB)=80°, m(∡BOC)=130°.

cazul 1 : daca AOB si BOC sunt adiacente.

m(∡AOB)+m(∡AOC)+m(∡BOC) =360° ⇔

m(∡AOC)=360°-(m(∡AOB)+m(∡BOC))=150°.

cazul 2 : daca AOB si AOC sunt adiacente.

m(∡AOB)+m(∡AOC)=m(∡BOC) ⇔ m(∡AOC)=m(∡BOC)-m(∡AOB)=

130°-80°=50°.

dar, avand in vedere a problema ne spune direct ca unghiurile sunt in jurul unui punct, al doilea caz cade, prin urmare doar primul caz convine.

2) b) (OD = bisect.∡AOB ⇔ m(∡AOD)=m(∡BOD)=1/2∡AOB=40°.

m(∡BOE)+m(∡EOC)=m(∡BOC) ⇒ m(∡BOE)=m(∡BOC)-m(∡EOC)=

130°-90°=40°.

m(∡DOE)=m(∡BOD)+m(∡BOE)=40°+40°=80°.