Question

Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 158 feet from camera 2, which was 121 feet from camera 3. Cameras 1 and 3 were 140 feet apart. What camera had to cover the greatest angle?

Answer 1

Camera 3 that has to cover at least `74.11^@`

Explanation:

The triangle of the figure below represents the case

Applying the Law of Cosines (`a^2=b^2+c^2-2ab*cos alpha`) we have:
`121^2=158^2+140^2-2*158*140*cos hat C 1` => `cos hat C 1 = (158^2+140^2-121^2)/(2*158*140)=(24964+19600-14641)/44240=29923/44240~=.67639` => `hat C 1= 47.44^@`

`140^2=158^2+121^2-2*158*121*cos hat C 2` => `cos hat C 2 = (158^2+121^2-140^2)/(2*158*121)=(24964+14641-19600)/38236=20005/38236~=.52320` => `hat C 2= 58.45^@`

`158^2=140^2+121^2-2*140*121*cos hat C 3` => `cos hat C 3 = (140^2+121^2-158^2)/(2*140*121)=(19600+14641-24964)/33880=9277/33880~=.27382` => `hat C 3= 74.11^@`

Since `hat C 3 > hat C 2 > hat C1`, the answer is Camera 3.

Note that `hat C 1+hat C 2 +hat C 3=47.44^@+58.45^@+74.11^@=180^@`, as it should be for internal angles of a triangle.