- #1
Whateverworks
- 20
- 0
Hello mathematicians!
I've recently completed a trigonometry course online and find the subject to be of great interest.
I find the laws of sine and cosine fascinating and extremely useful and also, of course, Pythagoras theorem is beautiful as well.
Firstly, I claim no superior knowledge here so if I say something wrong please correct me, that is why I am here.
Now to my question. As I have understood it, for one to make use of the laws of sine or cosine one needs to have some values to get started. I have a hard time figuring out if that is correct. I have used the last days trying to solve a problem I have given myself - however I seem lost and that may be because it's unsolvable!
If I know that a+b+c = x, and that a2+b2+c2=y, in a right triangle, what is the value of a, b and c, respectively.
For a specific case let's say, a+b+c = 70, and that a2+b2+c2=1682
Now if this is solvable I request that you do not post the answer by any means, I simply need to know if that can be solved and I will try again.
Thanks a lot,
//WeW.
I've recently completed a trigonometry course online and find the subject to be of great interest.
I find the laws of sine and cosine fascinating and extremely useful and also, of course, Pythagoras theorem is beautiful as well.
Firstly, I claim no superior knowledge here so if I say something wrong please correct me, that is why I am here.
Now to my question. As I have understood it, for one to make use of the laws of sine or cosine one needs to have some values to get started. I have a hard time figuring out if that is correct. I have used the last days trying to solve a problem I have given myself - however I seem lost and that may be because it's unsolvable!
If I know that a+b+c = x, and that a2+b2+c2=y, in a right triangle, what is the value of a, b and c, respectively.
For a specific case let's say, a+b+c = 70, and that a2+b2+c2=1682
Now if this is solvable I request that you do not post the answer by any means, I simply need to know if that can be solved and I will try again.
Thanks a lot,
//WeW.