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Math - trigonometri
Former Member
Posts: 1,876,323 The Mix Honorary Guru
How do I get about this excersize?
I have two different triangles, and get told that these facts abpply for both:
A = 50 degrees, a = 2,4 and b = 3
Now I need to calculate the rest of the sides and angles, but don't know how to, when there are two different solutions on how to do so.
I have two different triangles, and get told that these facts abpply for both:
A = 50 degrees, a = 2,4 and b = 3
Now I need to calculate the rest of the sides and angles, but don't know how to, when there are two different solutions on how to do so.
Post edited by JustV on
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Also referring to a and b; lengths right? Where from? a to b is 2,4 and b to c is 3?
CM? MM?
edit: meh, i can't remember how to do it correctly
And lower case letters refer to the length of the sides opposite that angle e.g. length a is referring to the side opposite the vertice labelled A. If I remember correctly...
It's a while since I've done any maths, let alone trigonometry. I'll give it a think over but I'm pretty sure there are others on here that are more capable of working it out and helping you.
Cos A = (b squared + c squared - a squared) / 2bc
I would know how to do this with the sinusrelation. But clearly that can't be valid, as I'd get one answer, and not two for the different triangles.
I still only get on answer from that.
Though should I see if I get different answers if trying one triangle with the cosine rule, and the other with the sine/sinus rules?
Thinking about it if two triangles have one angle and two lengths in common then surely the third length and the other two angles would also have to be the same for both triangles? The triangles in the picture don't look the same tho so maybe I've misunderstood the question.
Either waym since I am retarded as I am, then how do I solve the equation:
4-4x^2/(x^2+2x+1)^2=0
?
I hate going through old math :mad:
Maybe wheresmyplacebo can help you as he's doing a degree in Chemistry with Maths if I remember correctly. We go to the same uni but he's in a higher maths class than me!
Now, it's justthe fucking triangles doing my head in :mad:
I got a calculator, dont think I know what the hell your talking about like
I got ungraded in maths Hey seriously I did, was pathetic and still am
*sigh*
They key is in when you root the equation- the number can be two, of which the number on your calculator is only one.
Care to explain it again? Can't follow your thinking.
Thanks
I can't explain it very well, as I can't quite remember how to do it and I don't have my books to enhance my memory.
I think I mean that:
say x is 4. When you square root it root x can either be 2 or minus 2, because -2 X -2 = 4, or 2 X 2 = 4.
Soprry I'm not of more help, if I had my textbook I could have worked it out:(
About the triangle, you must have been "thought" of what cosA or sinA are equal to if there's no 90 degree on the triangle. I don't remember what it's equal to, but it should be in your book or something.
Gzatt, I used to rule at these things... Now I'm not even sure if I solved your equation right...
But a length cant be minus.
it can in quantum physics i think
4-4x²
= 0
(x²+2x+1)²
The solution to that is 1 and -1, got the answer with the help of my graph-calculator, whom I'd be dead without.
If the fraction is 0, then the thing below, (x²+2x+1)² has to be different than zero. For x=1, that is 16, so x=1 is acceptable. But for x=-1 it's equal to 0, and 0/0 zero can't work like that. So it's only x=1, unless I remember this less than I think.
Sorry
But thanks, it's appericiated.
We seem to do maths differently from you, Tal - we'd do very little of that graphically, despite the fact that I have a lovely graphics calculator.
The length = x, and the width(sp?) = y
x>y and the space is 15, while the circumference is 16.
Now, I can see that x=5, while y = 3, but how do I prove it with calculations?
And also, I know that the root of the equation x^2-6x+5 = 0 is x=1 and x =5.
But how do I find the definition-set (don't know if that's the word in English, I am guessing) of ln(x^2-6x+5 ) ?
x²-6x+5=0 => x=1 or x=5 , as you said.
Let's say that A=x²-6x+5 .
Trying a number lower than 1, for example 0, A=5. So:
For x<1, A>0.
For a number between 1 and 5, for example 2, A=-3. So:
For 1<x<5, A<0.
For a number greater than 5, for example 10, A=45. So:
For x>5, A>0.
So for A to be greater than zero, x has to be lower than 1 or greater than 5. The field of definition for ln(x²-6x+5) is (-œ, 1)u(5, +œ ) .
I used œ as the symbol for infinite (there's no such symbol in this font) and u for "connection".
However, I have some things to say:
1)The way you present a field of definition is possibly different that the way we do, so don't just copy these things over. Also, you might not have been taught about infinite, but I know of no other way to describe a field of definition.
2)Which grade are you in? This is like what we did in 9th, but you must be older than that.
3)(And by far most important) Coming to a forum and asking for answers to your problems isn't going to help you. Don't you have teachers and books over there? I'm sure you do. You're supposed to listen to your teachers, ask them questions and read your books so that you will UNDERSTAND how to find the solutions on your own. Asking for a little help isn't that wrong, but if you're going to succeed at your tests you must know how to find a solution to given problems. For example, in your book there's supposed to be proof of why what I said for this answer is correct, and you might be asked to prove it yourself. Studying isn't finding answers in any possible way, but learning what you're supposed to learn. As for the solution to the square (which can't be a square, or it would be x=y) the answer is so ridiculously easy that you could just open your book and read it.
Sorry for the lecture, but you must learn to do things yourself.