- 02 Apr '09 23:41Given: An isosceles triangle ABC with angles 80 / 20 / 80.

From A draw a straight line at 40 degrees from base to the opposite side. From C draw a straight line at 30 degrees from base to the opposite side.

Connect these opposite side points with a straight line.

Now solve for all interior angles without using trigonometric functions or a protractor. Then prove your answer is correct.

[if you think you can do this with simple algebra, you are mistaken] - 13 Apr '09 21:50 / 1 editAssuming A and C are points on the base, so that angle ABC=20 deg.

Point D is where the 40 deg line above the base AC intersects side BC. Point E is where the 30 deg line above the base AC intersects side AB. Point F is where the lines AD and CE intersect in the mid-region of the triangle ABC.

All angles are easy to find except BED, BDE, FDE and FED. By using the law of sines [ Example: sin(CED)/CD = sin(DCE)/DE] you can determine through algebra that

angle FED = sin^-1 [ {sin(DCE)*sin(CFD)} / {sin(EAF)*[ {1 - (sin(BCE)/sin(BEC))} / {1 - (sin(BAD)/sin(BDA)} ] + sin(FDC) } ]

Apologies for the numerous parentheses. This calculation gives an angle of 32.8 degrees for FED. The other angles are simple once this value is obtained. I'm sure there are other ways to use the law of sines to come up with one of the unknown angles, which may quite possibly be simpler.

Of course, after working through all that, I notice "without trig functions". - 14 Apr '09 07:39

I can only find bounds without using trig. Which is annoying...*Originally posted by sven1000***Assuming A and C are points on the base, so that angle ABC=20 deg.**

Point D is where the 40 deg line above the base AC intersects side BC. Point E is where the 30 deg line above the base AC intersects side AB. Point F is where the lines AD and CE intersect in the mid-region of the triangle ABC.

All angles are easy to find except BED, BDE, FDE and FED ...[text shortened]... be simpler.

Of course, after working through all that, I notice "without trig functions". - 25 Apr '09 01:09i'm pretty sure this is one of those situations where the original poster made a faulty assumption about what happens to the angles when you construct a new piece onto the triangle... the law of sines shows that (as sven1000 posted) angle FED is about 32.8 degrees, which is not obtainable by anything other than a reverse trig function.

as other posters have mentioned, one can construct bounds on angle FED (for example FED < 70, which is trivial to see) but without making a faulty assumption of congruency to another angle, or parallelism where there isn't parallelism, or bisection where there isn't bisection, the value of angle FED can't be nailed down without trig.

if the OP can contradict this by posting their solution (preferably an elegant extension of AE and CD that construct a new triangle that gives us insight into the value of FED?), i will humbly eat my hat and shower them with sincere awe and praise! cheers! - 25 Apr '09 12:58

...but can he perhaps wait maybe another week, as I would really like to try this again, but my dissertation is due in on Friday and, well, yes....*Originally posted by Aetherael***...if the OP can contradict this by posting their solution (preferably an elegant extension of AE and CD that construct a new triangle that gives us insight into the value of FED?), i will humbly eat my hat and shower them with sincere awe and praise! cheers!** - 01 May '09 23:12 / 1 edit

I drew a sketch of this one out, and I don't see any two triangles which are provably similar. As a matter of fact, since the large triangle is symmetrical, but the two angled lines are not, the third line does not run parallel to the base, which suggests to me none of the triangles are in fact similar.*Originally posted by camilli***Hint: think out of the box , and apply what you know about similar triangles.**

Not sure what you mean by drawing a box on the triangle either, to be honest.

And if someone using reverse trig functions determines the answer as an irrational value when all given angles are multiples of 10, that would indicate to me a lack of non-trig alternatives. - 02 May '09 08:23 / 1 edit

this is exactly what i was getting at in my post... and not to be offensive, but i think uzless did not get the essence of the difficulty of the problem - a rectangle constructed around the triangle gives us no further insight into the measures of the interior angles, which are the only angles whose measures are difficult to determine based on the confines of the problem statement.*Originally posted by geepamoogle***I drew a sketch of this one out, and I don't see any two triangles which are provably similar. As a matter of fact, since the large triangle is symmetrical, but the two angled lines are not, the third line does not run parallel to the base, which suggests to me none of the triangles are in fact similar.**

Not sure what you mean by drawing a box on the ...[text shortened]... ll given angles are multiples of 10, that would indicate to me a lack of non-trig alternatives.

and, as you said, since an "inelegant" (trigonometric-based) method for finding the measure of these angles shows them to have*irrational*measure, i find it very difficult to believe that there is some "non-trig" method for finding these values without an illogical and untrue assumption about the diagram that would "prove" these angles to have measures that they, in fact, do not.

of course, i would be happy to be proven wrong. but as a result of these statements (in the absence of further evidence or some assertion of proof that could be refuted), i personally think this problem is not worthy of further undue exploration. now the burden of proof is on the original poster! please show us your solution, or we will be perpetually despondent! - 05 May '09 09:05 / 1 edit

in the language of sven1000:*Originally posted by camilli***Given: An isosceles triangle ABC with angles 80 / 20 / 80.**

From A draw a straight line at 40 degrees from base to the opposite side. From C draw a straight line at 30 degrees from base to the opposite side.

Connect these opposite side points with a straight line.

Now solve for all interior angles without using trigonometric functions or a protra ...[text shortened]... your answer is correct.

[if you think you can do this with simple algebra, you are mistaken]

I've drawn round the triangle with a box, etc, to get the following angles

BED: 50

BDE: 100

and

DEF: 60

EDF: 10

Now, to work out why these differ from the trig solution... - 05 May '09 09:37

Something's wrong, because BED + BDE + DBE = 50 + 100 + 20 = 170.*Originally posted by Swlabr***in the language of sven1000:**

I've drawn round the triangle with a box, etc, to get the following angles

BED: 50

BDE: 100

and

DEF: 60

EDF: 10

Now, to work out why these differ from the trig solution...

Can you explain what you mean by drawing round the triangle with a box, etc? - 05 May '09 12:36 / 1 edit

Sorry - that should be a 110.*Originally posted by Palynka***Something's wrong, because BED + BDE + DBE = 50 + 100 + 20 = 170.**

Can you explain what you mean by drawing round the triangle with a box, etc?

To draw the box - extend the lines DE and AC and call the point where these lines intersect X. Then connect B, C and X into a box (the bottom lines of which should be XC).

Extending the line EC to the side of the box and call their intersection Y. We see that there are two triangles that are similar with corners at E. Clearly, CYX is 60 degrees.

Unfortunately, earlier I was under the impression that the other angle of this triangle on the vertical line was 60 degrees - not CYW - which is, of course, unfounded.

Thus, my solution is incorrect.

But -erm- that's how you draw the box... - 14 May '09 23:00De ja vu... I remember a puzzle like this one on this board from a long time ago. It was an equilateral triangle with an altitude and two other lines drawn. One angle was given (20 degrees), and the angle to solve for was ~32.54 degrees (I can't believe I still remember that number). I solved it with trig, but it took a while to convince some of the posters there that it had no simple solution (several people kept thinking it was 30 degrees). Does anyone remember what I'm talking about?