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Jacob's Geometry Ch. 3, Lesson 6 Question


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Guest tsmoffett

We may never get through Geometry at this rate.

 

Problem #11-15 in Set I are about the Sun Directions.

 

Can someone explain how the answer is 80 degrees for #12? How are we supposed to know that each angle is 40 or 50 degrees and not evenly divided to be 45 degrees each? Did we miss a basic instruction/reason somewhere?

 

Utterly lost, and using the DVDs to teach Geometry to my son. I had to do extra credit to get a C in the last semester of highschool Geometry....

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I don't have Jacob's (anymore); but we used Chalkdust. If you're using the "Ask Dr. Callahan" DVD's, his instruction isn't as good as Dana Moseley's on Chalkdust, although the Jacob's book (though somewhat wordy) sometimes gives better explanations than the Larson book used by Chalkdust.

 

However, I think if you can post the question in its entirety, some of us (perhaps not me) can help you! We are blessed to have some math geniuses on these high school boards, including Jane in NC, Jann in TX, LoriM, and others. I haven't seen Myrtle here in a while, but there are plenty here who can help you.

 

If the problem isn't too complicated, can you type it out for us?

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(I got such help with my Alg. 2 question here, I feel good about being able to pay it forward to help here.)

 

The chapter is about linear pairs and vertical angles. I tell my son, who's doing this book, to try to get into the author's head and figure out how he's trying to teach the topic of each chapter by forming questions.

 

I wish I could give a drawing here, but I can't, so here goes an explanation with words. Please turn in your hymnal to page 113...

 

I think one of the key words is "bisect" in the paragraph that starts out "sun directions." The lines of the map directions bisect the angles formed by these lines. The definition of "bisect" is on p. 99 -- "a line bisects and angle iff it divides the angle into 2 equal angles." The map direction lines bisect the sun direction lines (and not the other way around).

 

We are given NOA = 50.

Line NO bisects angle COA, so angle CON is also 50 degrees.

(this answers #11.)

Line EO bisects angle AOB, so AOE and EOB are equal.

We know what angle AOE equals -- If NOA is 50 degrees, and we assume the direction points are 90 degrees (NOE= 90), so AOE = 90-50, or 40.

Since (as I said), Line EO bisects angle AOB, and AOE and EOB are equal, then EOB also equals 40 degrees.

AOE=40, EOB=40, so AOB=80.

 

The answer book says the reason it's 80 degrees is because 130-50=80. I didn't get there that way! He's using the theorem on p. 111, that "the angles in a linear pair are supplementary." You can get there from #11, if you know all the angles. But finding the degrees of the angles is the first step. The bisection thing is what tells you what angle EOB and SOB are.

 

I made a sheet for my son (and myself) that has all the definitions, postulates, and theorems on it for each chapter. There's something similar in the back of the book at page 741, but that one doesn't include the definitions. I've written it out up to chapter 10. If you're interested I can send it to you. I wrote it out because I kept forgetting the previous chapters' theorems, and I hated to keep turning to the back of the book or hunting down where I thought I saw a definition. I think it was in chapter 3 that I decided to do that. :)

 

I can't answer algebra 2 questions, but I am enjoying (and understanding) this Jacobs geometry pretty well.

Edited by Laura K (NC)
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