Geometry question?

Question:First of all, I would like to know how to solve this problem, and not just what the answer isI also don't know how to do the sybols so I'll just explain it...Thanks for your help! Here is the problem:

Ray OC bisects the angle AOB, Ray OD bisects angle AOC, ray OE bisects angle AOD , Ray OF bisects angle AOE, adn Ray OG bisects angel FOC


a. if measure of angle BOF=120 then the measure of angle DOE=?

b. if the measure of angle COG=35 then the measure of angle EOG=?

Answers:
It helps to draw a picture of what they are telling you...

Draw angle AOB. and let "x" represent that angle of <AOB
(sorry... but I don't know how to do symbols either... so the "less than" symbol is to mean "angle" AOB)

Okay... Now bisect that angle... From vertex O, draw a ray and put a C at the end of the line you drew from O...

That means that < AOC and < COB = 1/2 < AOB = (1/2) x

Now if you split that in half again... that means that < AOD and < DOC = 1/2 < AOC = 1/4 <AOB = (1/4) x

Now if you split that in half again... that means that < AOE and < EOC = 1/2 < AOD = 1/4 < AOC = 1/8 < AOB = (1/8) x

Now if you split that in half again... that means that < AOF and < FOC = 1/2 < AOE = 1/4 < AOD = 1/8 < AOC = 1/16 < AOB = (1/16) x

Now that you have found all that... let's look at the first question...

A) If measure of < BOF=120, then the measure of < DOE=?

Okay... we just said that "x" is going to represent < AOB. and we found out that <AOF = (1/16) x , right?

What do we also know?

We know that < AOF + < FOB = < AOB . which then tells us that < AOB - < AOF = < FOB

Soo. x - (1/16) x = < FOB

That means that < FOB = (15/16) x

If < BOF = 120 degrees, what is x?

(15/16) x = 120 degrees

x = 120 (16/15) = 128 degrees . so < AOB = 128 degrees
... with me?

That means that < AOC = 1/2 < AOB = 1/2 (128 deg) = 64 deg

We also know that <AOF = < AOB - < FOB = 128 deg - 120 = 8 degrees

We also know that <AOF = 1/2 < AOE . so because <AOF = 8 degrees... that means that < AOE = 16 degrees

We know that < AOE = < DOE because the problem tells us that ray OE bisects < AOD.

< AOE = 16 degrees. so that means that < DOE = 16 degrees

< DOE = 16 degrees ** ANSWER **
____________________________

Now ... let's look at the second question...

B) If the measure of < COG=35, then the measure of < EOG=?

Okay... scratch the angles that we found in Part A) (you know, the bit about how we found that < AOB = 128 degrees from know that < FOB = 120 degrees), but "remember" all the relationships that we found before we started answering Part A)

Remember, we said that "x" is going to represent < AOB. and we know that < AOC = 1/2 < AOB = (1/2) x

We also found out that <AOF = (1/16) x , right?

What do we also know?

We know that < FOG = < GOC because the problem tells us that ray OG bisects < FOC.

We know that < AOF + < FOG + < GOC = < AOC .

When we substitute the relationships that we know in terms of "x"... and < COG = 35 degrees... which also means that < FOG = 35 degrees... we get...

(1/16) x + 35 degs + 35 degs = (1/2) x

** Now solve for "x".**

(1/16) x + 70 degs = (1/2) x

(1/2) x - (1/16) x = 70 degs

(8/16) x - (1/16) x = 70 degs

(7/16) x = 70 degs

x = (70 degs)(16/7) = 160 degrees

So because x = 160 degs... we know now that < AOB = 160 degs

Now we have to set up another equation so that we can find < EOG. so let's write one which shows what angles PLUS < EOG will make up < AOC. what do you see on your drawing?

You should see this...

We know that < EOG = < FOG - < FOE .

When we substitute the relationships that we know in terms of "x"... and we know that < FOG = < COG = 35 degrees... we get...

< EOG = 35 degs - (1/16) x

Remember we found that x = 160 degs in this Part B) of the question...?

So, since x = 160 degs... just plug it in for "x"... like this...

< EOG = 35 degs - (1/16)(160 degs) = 35 degs - 10 deg = 25 degs.

< EOG = 25 degrees ** ANSWER **

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