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?
Camera 1
None
Camera 3
Camera 2 **
or camera 3 I'm not sure
To determine which camera would cover the greatest angle, we need to consider the distances between the cameras and apply the concept of trigonometry.
Let's label the corners of the triangular parking lot as A, B, and C, with camera 1 at corner A, camera 2 at corner B, and camera 3 at corner C.
We know that camera 1 is 110 ft away from camera 2 (the distance between A and B), and camera 2 is 137 ft away from camera 3 (the distance between B and C). Additionally, camera 1 and camera 3 are 158 ft apart (the distance between A and C).
Now, we can use the Law of Cosines to find the angles at each corner of the triangle. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C (opposite side c):
c^2 = a^2 + b^2 - 2ab * cos(C)
Let's find the angle at corner A (opposite side a):
a = 110 ft
b = 158 ft
c = 137 ft
c^2 = a^2 + b^2 - 2ab * cos(C)
137^2 = 110^2 + 158^2 - 2(110)(158) * cos(C)
18769 = 12100 + 24964 - 2(110)(158) * cos(C)
18769 = 37064 - 34960 * cos(C)
-18295 = -34960 * cos(C)
Now, we can solve for cos(C):
cos(C) = -18295 / -34960
cos(C) ≈ 0.52277
To find the angle C, we can use the arccosine function (cos^(-1)):
C ≈ cos^(-1)(0.52277)
C ≈ 59.92 degrees
Similarly, we can find the angle at corner B (opposite side b):
a = 137 ft
b = 158 ft
c = 110 ft
Let's find the angle at corner B (opposite side b):
c^2 = a^2 + b^2 - 2ab * cos(B)
110^2 = 137^2 + 158^2 - 2(137)(158) * cos(B)
12100 = 18769 + 24964 - 2(137)(158) * cos(B)
12100 = 43733 - 43348 * cos(B)
-31633 = -43348 * cos(B)
Now, we can solve for cos(B):
cos(B) = -31633 / -43348
cos(B) ≈ 0.72975
To find the angle B, we can use the arccosine function:
B ≈ cos^(-1)(0.72975)
B ≈ 43.08 degrees
Finally, to find the angle at corner C (opposite side c):
a = 158 ft
b = 110 ft
c = 137 ft
c^2 = a^2 + b^2 - 2ab * cos(A)
158^2 = 110^2 + 137^2 - 2(110)(137) * cos(A)
24964 = 12100 + 18769 - 2(110)(137) * cos(A)
24964 = 30869 - 29960 * cos(A)
-5905 = -29960 * cos(A)
Now, we can solve for cos(A):
cos(A) = -5905 / -29960
cos(A) ≈ 0.19688
To find the angle A, we can use the arccosine function:
A ≈ cos^(-1)(0.19688)
A ≈ 78.99 degrees
Based on the angles that we have calculated, we can see that camera 2, which is at corner B, would cover the greatest angle with approximately 43.08 degrees.
direct application of the cosine law,
using the fact that the largest angle is opposite the largest side ....
158^2 = 110^2 + 137^2 - 2(110)(137)cosØ
30140 cosØ = 5905
cosØ = .195919...
Ø = appr 78.7°
Answer accordingly.