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Geometry Lines and Angles.

+2 votes
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!!
asked Dec 25, 2012 in GEOMETRY by 1Dfangirl14 Rookie

2 Answers

+2 votes

Draw a picture for current situation

geometry homework help

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

answered Dec 26, 2012 by steve Scholar
+1 vote

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

geometry_lines_angles

answered Dec 26, 2012 by Naren Answers Apprentice

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