Question

Find m∠B, given a = 11, b = 12, and c = 17.

I used the law of cosines to get cos(b) = (11^2 - 12^2 + 17^2) / (2 * 11 * 17) which is equal to 40.75 degrees but that is not in my answer list. These are the ones I can choose:

m∠B = 49.9º
m∠B = 40.1º
m∠B = 45.3º
m∠B = 44.7º

Answer 1

Oops. Wrong answer. Let me repost it.

Answer 2

Here: cos(b) = (11^2 - 12^2 + 17^2) / (2 * 11 * 17)

Answer 3

the convention is angle B is opposite side b.
the law of cosines for this problem is
b^2 = a^2 + c^2 - 2 a c cos(B)
12^2 = 11^2 + 17^2 - 2*11*17*cos(B)
now solve for cos(B)
12^2 - 11^2 -17^2 = -2*11*17*cos(B)
cos(B)= (11^2 +17^2-12^2)/(2*11*17)

that looks like your formula. what did you get for an answer ?

Answer 4

cos(B) = 0.711
B= acos(0.711)

type into google
acos(0.711) in degrees=

Answer 5

But by the information on this page ( http://mathworld.wolfram.com/LawofCosines.html) the minus sign should be in front of the b^2. That is what's confusing me.

Answer 6

So shouldn't it be cos(B) = (11^2 - 12^2 + 17^2) / (2 * 11 * 17)

Answer 7

Ah. I was typing the question into Wolfram Alpha and it said B was equal to 40.75 degrees. Why didn't it solve it correctly?

Answer 8

you have
12^2 = 11^2 + 17^2 - 2*11*17*cos(B)

you could write this as
12^2 - 11^2 - 17^2 = -2*11*17*cos(B)
and
cos(B)= (12^2 - 11^2 - 17^2)/(-2*11*17)

or you could start with
12^2 = 11^2 + 17^2 - 2*11*17*cos(B)
add 2*11*17*cos(B) to both sides:
2*11*17*cos(B) + 12^2 = 11^2 + 17^2
and
2*11*17*cos(B) = (11^2 + 17^2 - 12^2)
and
cos(B) = (11^2 + 17^2 - 12^2)/(2*11*17)

which is the same as the first expression, except top and bottom have been multiplied by -1

both expressions give the same answer.

Answer 9

I got the correct answer now which is 44.68 degrees.