Originally posted by sven1000I can only find bounds without using trig. Which is annoying...
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".
Originally posted by Aetherael...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....
...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!
Originally posted by camilliI 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.
Hint: think out of the box , and apply what you know about similar triangles.
Originally posted by geepamooglethis 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.
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.
Originally posted by camilliin the language of sven1000:
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]
Originally posted by SwlabrSomething's wrong, because BED + BDE + DBE = 50 + 100 + 20 = 170.
in the language of sven1000:
I've drawn round the triangle with a box, etc, to get the following angles
Now, to work out why these differ from the trig solution...
Originally posted by PalynkaSorry - that should be a 110.
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?