15. Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 110 ft from camera 2, which was 137 ft from camera 3. Cameras 1 and 3 were 158 ft apart. Which camera had to cover the greatest angle? (1 point)
camera 1
There is not enough information to tell.
camera 3
camera 2

Question

15.   Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 110 ft from camera 2, which was 137 ft from camera 3. Cameras 1 and 3 were 158 ft apart. Which camera had to cover the greatest angle?  (1 point)
camera 1
There is not enough information to tell.
camera 3
camera 2

anonymous · Answer

um...

anonymous · Answer

You have a triangle with side lengths of 110, 137 and 158. And you need to find the angles in between them.  Do you know how to use law of cosines and law of sines?

anonymous · Answer

no

anonymous · Answer

Well the law of cosines goes like this c^2 = a^2 + b^2 -a*b*cos(C) where a, b, and c are side lengths and C is an angle.

anonymous · Answer

.-.

anonymous · Answer

To get the angle opposite C you would need to do C = arccos((a^2+b^2-c^2)/ab)
I used algebra to get that answer.
You can decide what a b and c are that relate to 110 137 and 158
(sorry if this doesn't make much sense =/ I'm not very good at explaining.)

anonymous · Answer

if a = 110, b = 137, and c = 158 then you would have the angel C = arccos((110^2+137^2-158^2)/(110*137))