I've got an equation that goes like this: \(k=\tan{\left[\frac{\arctan{\left(\frac{a}{b}\right)}-\arctan{\left(\frac{c}{d}\right)}}{2}\right]}\) where a, b, c, d are all positive integers. How do I find an exact value of k? I'm trying to find the gradient of an angle bisector.
Basic trigonometry hint: use tan(a-b)=.... followed by tan(a/2)=.... If you do it right, you should end up with no trigonometric functions. If you do it wrong.... Huh?
Is that a really subtle innuendo? That's just so unnecessary and cruel man. Please Register or Log in to view the hidden image! *haiz* Most people would stop writing after "...trigonometric functions." I really appreciated you responding so quickly, now you've unnecessarily lessened it. lol.
I was wondering if you read this somewhere or if you made it up all by yourself , because it sounds like pure BS What is the "gradient of an angle bisector"? So, what's the answer?
Well I guess you are prone to such things. But I'm cool with it. Trying to express an angle bisector in Cartesian coordinates. It's part of a geometrical problem that needs to be solved using Cartesian graphs. \(\frac{ ac + bd - \sqrt{(a^2 + b^2)(c^2 + d^2)} }{ bc - ad }\)
The gradient of a line in Cartesian form, which represents an angle bisector. Don't worry, it's the same. Anyway I've already manged to solve my problem. So thanks. Please Register or Log in to view the hidden image!
"Gradient" of a line? You mean the slope of a line. If you want to find out the slope of the bisector line when you know the slopes of the two lines forming the angle, your starting formula for \(k\) is all wrong so, the whole exercise has been a waste of time. See if you can get the correct formula for \(k\). Otherwise , what was the "gradient of the angle bisector" you were talking about? I pretty much doubt it, you just copied the answer, you don't know how it was derived. Based on this and on the fact that your starting point is all wrong, I can see that you are going to get an F on your trigonometry homework. I told you to pay attention to what you are doing, otherwise you'll get it all wrong but you had to be a smartalleck about it.... .
No, my starting is correct. First of all, if you want me to call me a smart aleck at least spell it correctly. Please Register or Log in to view the hidden image! It is right and I was attentive while working out the sums at my desk. But I already did it and solved my problem. LOL I would explain to you why the starting is correct but I've already finished.
You sure about it? Because it isn't. You need to learn basic trigonometry. Show your derivation and I'll show you where you made your mistake. Too bad that you already turned your homework in, you'll get an F, I can guarantee that. This should teach you not to be so arrogant.
I'm not being arrogant, it really is correct. Please Register or Log in to view the hidden image! I haven't fully explained the question to you but I've already completed it. Anyway your post #2 helped a lot.
If you are trying to find out the slope of the bisector, as I think you do, then your formula is incorrect. Publish your derivation and I'll show you where you made your error. You are going to have to put that tongue back in your mouth. I know it did, unfortunately you have the wrong final answer to your homework. Next time , do a better job at explaining the problem statement and at showing your derivation and you'll avoid the embarrassment of getting the wrong answer.
Don't be too disappointed when your homework comes back graded "F". Hopefully, this should be a valuable lesson.
A lesson about arrogance? I did tell you I was done, thanks for your concern but it was a bit excessive. This thread should've ended by #2 or #3, so let's just end it here.
It couldn't, since your formula in post 1 is wrong. You don't get it, do you? When the homework comes back, be honest and post your grade.
