Lib. Arts Urgent call-out to Math Nerds!!!

[MistyJ]

Saxon 8/7, Lesson 68, #20 page 469.

 

Angle ADC is 60 degrees. So, the opposite, angle CBA is 60 degrees.

DCB is 120. DAB is 120. For the life of me, I can't figure out DAC or CAB.

 

Help

[mcconnellboys]

I don't have this book, but are you describing a parallelogram? In case you are, here's a link to the rules for solving those type angles: http://www.mathwarehouse.com/geometry/quadrilaterals/parallelograms/

 

I hope that helps. I tried to draw out what you described, but I'm just not sure I'm doing it correctly at all, sorry....

[kalanamak]

Off the top of my head (with no Saxon book) I say the angle you are describing cuts the 120 in half. It is a long diamond with a line between the two points that are closest together, right?

 

Half of 120 is 60, and the angles of the two triangles that you have created with the line in the middle have three 60 degree angles.

 

Or am I seeing this all wrong?

[MistyJ]

Off the top of my head (with no Saxon book) I say the angle you are describing cuts the 120 in half. It is a long diamond with a line between the two points that are closest together, right?

 

Half of 120 is 60, and the angles of the two triangles that you have created with the line in the middle have three 60 degree angles.

 

Or am I seeing this all wrong?

 

 

That's what I thought, but the answer key says different.

[kalanamak]

That's what I thought, but the answer key says different.

 

typo? If one angle is 60 and the other two equal each other, how can they not all be 60?

[Matryoshka]

Well, I first thought this problem was easy until I sketched it and realized it had to involve the diagonal. The problem is that the bisected angles aren't congruent to any of the angles given. It seems intuitive that the diagonal would bisect the 120º angle, but, as I thought about it, unless it's a rectangle, I don't think it's possible that it would - the more the parallellogram tilts, the more unequal the two angles formed by the bisection of the diagonal would have to be.

 

So I drew the blasted thing with a protractor and ruler to clarify my thinking. The answer appears to be that one angle is 40º and the other 80º (as I measured it). But I still can't think what the geometric proof reason is - neither angle is opposite or adjacent to the horizontal line. We only know one of the three angles in each triangle the diagonal makes, so that isn't helping. Is there a piece of missing information? I feel like I must be missing something obvious...

[AngieW in Texas]

Angles DAC and ACB have to be equal to each other because they are alternate angles on a transversal between parallel lines.

 

Angles BAC and ACD have to be equal for the same reason.

 

Triangles DAC and BCA are congruent by ASA because angles ACB and CAD are equal, side AC is equal to itself, and angles DCA and BAC are equal.

 

I don't see how you can get an actual angle measure for DAC or CAB without knowing some other information. If you know the lengths of the sides, then the angle sizes will be proportional and you can figure them out that way since you know they have to add up to 120. If you know any of the angles DAC, ACD, BAC, or ACB, then you can figure out all the others.

 

I don't see how to figure out the angles with just the information you have posted.