Triangle Puzzle

Originally posted by TDR1
i think this is true.....opposite angles in a trapazoid are supplementary.....im not 100% sure bout this....but with this you can conclude that the answer is 60....im not taking random guesses...that may be wat it seems like

Opposite angles are supplementary if two of the sides are parallel, which we do not know is the case with trapazoid BCED; I think we can more or less rule that out, because the base angles are the same, but the points D and E are definitely on different parts of their respective lines, given the angles they form with vertices C and B respectively.

Nemesio

10 Degrees!

Is this right, wheres the original poster gone?

Why do you think it is right? Show the math, because I don't see it.

Originally posted by nemesio
Opposite angles are supplementary if two of the sides are parallel, which we do not know is the case with trapazoid BCED; I think we can more or less rule that out, because the base angles are the same, but the points D and E are definitely on different parts of their respective lines, given the angles they form with vertices C and B respectively.

Nemesio

thats true....my bad....i cant seem to get this

OK I'm assuming you've got all angles calculated apart from ADE, AED, CDE and BED.

We know that AB = AC so we assign a distance lets say 10cm (it does not matter) to these sides.
We also know that ABC = ACB so we can just use a simple ratio to find out where on AB point D is, also where on AC point E is. We can do this because we know that BCD = 60 degrees which is 75% of the ACB angle (80 degrees) meaning point D is 75% on the way along AB. (use same method for point E)
Using this method we now know that AD = 2.5cm and AE = 3.75cm, we already have the BAC angle (20 degrees) so with these 3 bits of info its now just basic triangle math.......

Finished a bit short there...

.......to calculate angles ADE and AED, and we all know that angles on a straight line = 180 degrees so CDE can now be easily obtained.

So I still think CDE = 10 degrees?

Originally posted by timebombted
[snip] We can do this because we know that BCD = 60 degrees which is 75% of the ACB angle (80 degrees) meaning point D is 75% on the way along AB.

This reasoning flawed. A ratioed division of one of the two identical angles in an isoceles triangle does not necessarily result in a comparable ratio with the divided line opposite the angle.

Take for example an isoceles triangle MNO, where aM is 30, and aN and aO are 75. MN=MO=10 units. Now, draw a line from N onto line MO at point P and make it a 90 angle. Because triangles always total 180, aMNP is 60 and aONP is 15. This makes a 60/75 ratio or, by your math, divides the line into MP=8 and PO=2.

However, triangle MNP is a 30/60/90 triangle, giving precise values for the lines. Since MN is 10, then NP is 5, and MP is 5 rad3 (square route of 3), or 8.66025404.

You are not applying sound geometrical math. Sorry.

TDR et al.

Of course a trapazoid has two parallel lines! BCED is not a trapazoid, it simply a quadralateral. I got my terminology screwed up somehow; and yes, opposite angles are complimentary on a trapazoid--it's just that this isn't one!

Sorry, I didn't catch that when I wrote above.

Originally posted by nemesio
This reasoning flawed. A ratioed division of one of the two identical angles in an isoceles triangle does not necessarily result in a comparable ratio with the divided line opposite the angle.

Take for example an isoceles triangle MNO, where aM is 30, and aN and aO are 75. MN=MO=10 units. Now, draw a line from N onto line MO at point P and make it a 90 ...[text shortened]... re route of 3), or 8.66025404.

You are not applying sound geometrical math. Sorry.

HINT: I think you need to use the law of sines to solve this:

For any triangle JKL: sin(<JKL)/JL = sin(<LJK)/KL = sin(<KLJ)/JK

Originally posted by richjohnson
HINT: I think you need to use the law of sines to solve this:

For any triangle JKL: sin(<JKL)/JL = sin(<LJK)/KL = sin(<KLJ)/JK

Do you really need trig here? I have gotten to a point where the angle must be no more the 50 degrees with various geometric rules...do I have to dust off my math book here?

Originally posted by nemesio
Do you really need trig here? I have gotten to a point where the angle must be no more the 50 degrees with various geometric rules...do I have to dust off my math book here?

You could always find a protractor and a ruler...

trig is used when u have at least one and one side or two sides...we cannot use trig here

Yeah. I was just reviewing the laws of sines and cosines, and all require some concrete side knowledge. We know a few things about some of the sides because of the isoceles triangles that are made, as well as a few 30/60/90 triangles, but none that contribute meaningfully to solving the problem. I think the &quot;Laws of Sines&quot; post was a red herring, but I enjoyed revisiting the math.

I certainly enjoyed it more than I did math class in high school, actually. Weird.

Nemesio