Question

Three security cameras were mounted at the corners of a triangles 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

Answer 1

Camera 2nd has to cover the maximum angle, i.e.
`78.70^\circ`.

Explanation:

Please have a look at the triangular park represented as a triangle
`\triangle ABC` with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles
`\angle A, \angle B, \angle C`

As per Law of cosine:

`cos C = (a^(2)+b^2-c^2 )/(2ab)\\cos B = (a^(2)+c^2-b^2 )/(2ac)\\cos A = (b^(2)+c^2-a^2 )/(2bc)`

Putting the values of a,b and c to find out angles
`\angle A, \angle B, \angle C`.

`cos C = (110^(2)+158^2-137^2 )/(2* 110 * 158)\\\Rightarrow cos C = (12100+24964-18769 )/(24760)\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ`

`cos B = (110^(2)+137^2-158^2 )/(2* 110 * 137)\\\Rightarrow cos B = (12100+18769 -24964)/(30140)\\\Rightarrow cos B = (5905)/(30140)\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ`

`cos A = (158^(2)+137^2-110^2 )/(2* 158 * 137)\\\Rightarrow cos A = (24964+18769-12100)/(43292)\\\Rightarrow cos A = (31633)/(43292)\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ`

Camera 2nd has to cover the maximum angle, i.e.
`78.70^\circ`.