Triangle Problem

You're so hostile toward me lol. Fun is one thing, but I'm looking at efficiency. When I make Rubik's Cube threads asking for help, it's mostly about me getting the answer for credit in my class and/or for me to understand the/a way to do the problem that works. It has nothing to do with fun.

I brought this to Southperry because I found it fun and so others could enjoy the challenge. This was completely extracurricular and meant to be fun, so no, it's not all about efficiency.

I looked at the problem with a pen and paper type approach and it is very complicated, you need to label sides, and use the law of sin/cosine to find other angles/sides. It's not simple to solve at all that way.

And now you realize that it's not some trivial question like you said earlier.

Which is why I took the liberty of drawing the triangle out (on a ghetto piece of loose leaf without being that accurate *twiddles thumbs...*) and I got the angle to measure 32 degrees, I'm sure I'm off by some degrees so I'd stick with Rethik's response of 30 degrees.

And if you were to try that on a test in anything past grade 4, you would fail. The most efficient answer isn't always the right one.

Did not see your extracurricular thing. Lol. And I was worried to get you the points..Or I wouldn't have even drawn it..

If it was a real question I would have used graph paper and made a really accurate triangle, believe me.

You would still fail, have you really never done geometry? You're supposed to use math, not a ruler and protractor.

./done being attacked

Have more fun SP members who want to do this.

./done being attacked

Have more fun SP members who want to do this.

dunno if you'll see this, but he's on the right man, the whole purpose of this is to put a challenging problem to those who want to spend some time solving it for kicks, and for me, solving something using drawings simply doesn't work, it's not precise enough

he could have worded it more uhhh, passively i guess? but i gotta agree with the guy.

now, will someone show me the math to solve this please? <.<

dunno if you'll see this, but he's on the right man, the whole purpose of this is to put a challenging problem to those who want to spend some time solving it for kicks, and for me, solving something using drawings simply doesn't work, it's not precise enough

he could have worded it more uhhh, passively i guess? but i gotta agree with the guy.

now, will someone show me the math to solve this please? <.<

Essentially, you just use a combination of the Laws of Sine and Cosine for (tri)angles to find all the lengths needed to find theta.

I expect to see a more elegant approach from pie's instructor, though, seeing how this is a rather... bullfighting method.

Essentially, you just use a combination of the Laws of Sine and Cosine for (tri)angles to find all the lengths needed to find theta.

I expect to see a more elegant approach from pie's instructor, though, seeing how this is a rather... bullfighting method.

oh, i know that, i am just too lazy to do it (and can't remember well how to do it, so i'd probably make mistakes), hence why i want someone to do it instead, and i can see their math.

oh, i know that, i am just too lazy to do it (and can't remember well how to do it, so i'd probably make mistakes), hence why i want someone to do it instead, and i can see their math.

Refer to my earlier post:

 Spoiler

Setting lengths AB and AC to an arbitrary value of 1:

Given the answer of 30 degrees perhaps there's a way to work back to a non-trig method.

For example, ADE is congruent to CFE. This suggests 'unfolding' to make D'C and AB parallel.

Then rather than finding theta (it's obviously 30 if this diagram works) you just need to prove that D' is where AD', CD', and BD' intersect. Or alternatively that BE and ED' are collinear, if you drop the unknown angles.

Essentially, you just use a combination of the Laws of Sine and Cosine for (tri)angles to find all the lengths needed to find theta.

I expect to see a more elegant approach from pie's instructor, though, seeing how this is a rather... bullfighting method.

My teacher's solution is to draw line DF so that it is parallel to line BC, then connect F and B intersecting at point G. Then find isosceles and equilateral triangles (BGC and DFG and equilateral, CGE, BCE are isosceles) then prove that DEFG is a kite.