Yeah uh. I didn't figure out the WAY to do it, but it's probably cos theorem used in a puzzle-like way.
#figuring out an angle from what I presume is the cos theorem?
deft sorrel
limber yoke
Maybe multiplying a+b-c might help
deft sorrel
Multiplying how?
limber yoke
$(a + b + c )\color{blue}(a + b - c)\color{black} = \frac{a.b}{a + b - c} \color{blue}(a + b - c)$
hoary sierraBOT
Momentarily dumb
deft sorrel
$(a + b + c )\color{blue}(a + b - c)\color{black} = \frac{a.b}{a + b - c} \color...
Waitwaitwait.
How are those sides equal?
limber yoke
If you cancel out the blue thing on both sides you are left with what you have given
deft sorrel
If you cancel out the blue thing on both sides you are left with what you have g...
I mean like
how is a + b + c = ab/(a + b - c)
This doesn't seem particularly obvious
limber yoke
I mean like
how is a + b + c = ab/(a + b - c)
Well it isn't always true. It would be only true for specific sets of numbers.
Question already gives the equality.
Which mean a,b,c can't be just anything. They need to be from those specific sets of numbers.
deft sorrel
Oh, you rewrote it to be that (x+y)(x-y) style of thing
more specifically, (a+b)^2 - c^2 = ab
AAAAH
that cancels out! so you get ab!
deft sorrel
Where can we go from here?
static crater
Where can we go from here?
as you said, cos theorem
deft sorrel
How can we utilize it here?
Wait, let me see if I still remember cos theorem
c^2 = a^2 + b^2 - 2abcosc
we have c^2 = a^2 + b^2 + ab
so
ab must be equal to -2abcosc
so
cos C times -2 = 1
cos C is...-1/2?
am I on the right track?
static crater
yes
deft sorrel
and if cosC is -1/2, it must be... um...
cosinus had that rule where two angles with a sum of 180 were the - of each other
if cosx = 1/2, then cos(180-x) = -1/2
x here is 30, so 180-x is 150 and that is our answer.
?
deft sorrel
was it cos60
static crater
yes