The Law of Cosines ought to work. Perhaps you could write down what it is? Maybe you've got a typo in the equation.Originally Posted by helpmewithtrig
-Dan
i have 3 sides of a nonright triangle , length 5(a), 2(b) and 5.4(c), and i want angle C.
i have tried law of cosines over and over but keep getting anle of about 90deg, i know it cant be 90deg, because its not a right triangle.
i must be doing sumthing really stupid, but cant seem to firgue out what is wrong.
please help somem1?
thanks for the reply topsquark.
the equation i am using is:
c^2 = a^2 + b^2 - 2ab cos(C)
i rearrange to get:
cos(C) = (a^2 + b^2 - c^2) / -2ab
when i plug in the numbers i get:
Cos (C) = -0.008
and then,
acrccos (-.008) = 90.46deg
thanks for any insight.
a = 5
b = 2
c = 5.4
angle C = ???
I'm home now, so I finally have my precious calculator in my hands.
helpmewithtrig I'm not sure what your problem is. You did the problem correctly. You DON'T have a right triangle because angle C is not 90 degrees, but 90.4584 degrees. There's nothing wrong with this answer!
-Dan
Hi:
If c^2 = a^2 + b^2 - 2ab cos(C), then cos(C) = (a^2 + b^2 - c^2)/(2ab). You divided by -2ab rather than its positive counterpart. Appropriate substitution of side lengths puts m<(C) at 90.458° to three places.
Regards,
Rich B.
PS: m<(C) translates to "measure of..."