Geometry Lines and Angles.

Question

I\'m having a really tough time with a few Geometry problems on my homework. Can someone please explain to me how to solve this problem? Line OC bisects Angle AOB, Line OD bisects angle AOC, Line OE bisects angle AOD, Line OF bisects angle AOE, and Line OG bisects Angle FOC. If m angle BOF=120, then m angle DOE=? This is a tenth grade honors geometry problem. Please Help!!

Answer 1

Draw a picture for current situation

Draw angle AOB.... and let θ represent the angle ∠AOB

That means that ∠AOC and ∠COB = 1/2 ∠AOB = (1/2)θ 
Now, split that in half again.

∠AOD and ∠DOC = 1/2 ∠AOC = 1/4 ∠AOB = (1/4)θ

Split that in half again.

∠AOE and ∠EOC = 1/2 ∠AOD = 1/4 ∠AOC = 1/8 ∠AOB = (1/8)θ

Split that in half again.

∠AOF and ∠FOC = 1/2 ∠AOE = 1/4∠AOD = 1/8∠AOC = 1/16∠AOB = (1/16) θ

Given that ∠BOF=120

We know that ∠AOF + ∠FOB = ∠AOB

∠AOB - ∠AOF =∠FOB 

θ - (1/16)θ = ∠FOB

∠FOB = ( 16 - 1) /15 θ = (15/16) θ 

Substitute ∠BOF = 120 degrees

(15/16) θ = 120 degrees 

 θ = 120 (16/15) = 128 degrees

∠AOB = 128 degrees 

 ∠AOC = 1/2 ∠AOB = 1/2 (128 deg) = 64 deg 
∠AOF = ∠AOB - ∠FOB = 128 deg - 120 = 8 deg 

We know that ∠AOE = ∠ DOE because the ray OE bisects ∠AOD....

∠AOE = 16 degrees.... so  ∠DOE = 16 degrees 

∠DOE = 16 degrees

 

I hope it helps

Answer 2

Given that ∠BOF=120
From Figure
∠EOF + ∠DOE + ∠DOC + ∠BOC = 120°
β  + 2β  + 4β  +  8β = 120
15β = 120
β = 8°