# given 3 sides, want an angle

• Apr 8th 2006, 02:00 PM
helpmewithtrig
given 3 sides, want an angle
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.
• Apr 8th 2006, 02:52 PM
topsquark
Quote:

Originally Posted by helpmewithtrig
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.

The Law of Cosines ought to work. Perhaps you could write down what it is? Maybe you've got a typo in the equation.

-Dan
• Apr 8th 2006, 03:15 PM
helpmewithtrig
what i did

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 = ???
• Apr 8th 2006, 04:44 PM
ThePerfectHacker
Quote:

Originally Posted by helpmewithtrig
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.

I think it is easier to use Heron's Theorem for these problems. Do you what it is?
• Apr 8th 2006, 04:58 PM
helpmewithtrig
herons theorum, isn't that an area formula? thing is, it may be easier, but the cosine law should work right? but i always get 90deg as answer.
• Apr 8th 2006, 05:15 PM
ThePerfectHacker
Quote:

Originally Posted by helpmewithtrig
herons theorum, isn't that an area formula? thing is, it may be easier, but the cosine law should work right? but i always get 90deg as answer.

The cosine thing would also work just as Heron's Formula use whatever you like more.

It looks like 90 degress.
• Apr 9th 2006, 03:20 PM
helpmewithtrig
cosine law
any1 know why cosine law wont work???
thanks some1 out there.
• Apr 9th 2006, 03:49 PM
ThePerfectHacker
Quote:

Originally Posted by helpmewithtrig
any1 know why cosine law wont work???
thanks some1 out there.

It always does :)
• Apr 10th 2006, 03:46 AM
topsquark
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
• Apr 10th 2006, 07:11 AM
c_323_h
Quote:

Originally Posted by topsquark
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

math is the science (or art) of precision
• Apr 12th 2006, 10:15 AM
Rich B.
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..."