I'm working on a question and I've come up with this:
$\displaystyle a^2 = 400+c^2-40c*\cos A $
$\displaystyle 400=a^2+c^2 - 2ac * \cos 120$
$\displaystyle c^2=a^2+400-40a*\cos C $
Is this possible to do?
I'll take a swing!
For starters, cos120 = -0.5
Assuming that a and A are the same value, we get the following:
a^2 = 400 + c^2 - 40c * cos(a)
400 = a^2 + c^2 + ac
c^2 = a^2 + 400 - 40a * cos(c)
a^2 = 400 + a^2 + 400 - 40a * cos(c)
<=>
0 = 800 - 40a * cos(c)
<=>
0 = 800 - 40c * cos(a)
<=>
800 - 40a * cos(c) = 800 - 40c * cos(a)
<=>
40a * cos(c) = 40c * cos(a)
<=>
a * cos(c) = c * cos(a)
<=>
a = c * cos(a) / cos(c)
<=>
c = a * cos(c) / cos(a)
And that's as far as I got I'm probably forgetting some trig rules. Do you know anything else about these values?
All you have done is applied the cosine rule to a triangle with a side of 20 units and opposite angle 20 degrees, phrasing it three different ways. There is nowhere near enough information in what you have posted.
Post the whole question that you're working on please. Include ALL known information.
Hello, jgv115!
I'm working on a question and I've come up with this:
. . $\displaystyle \begin{array}{ccc} a^2 &=& 400+c^2-40c\cos A \\
400 &=& a^2+c^2 - 2ac\cos 120 \\
c^2&=&a^2+400-40a\cos C\end{array} $
Is this possible to do? . . . . no
We have a triangle with: .$\displaystyle B = 120^o,\;b = 20$
And (it seems) we are expected to determine the other sides and angles.
But there are brizillions of such triangles.
As mr. fantastic requested, please give us the original problem.