Question

Three security cameras were mounted at the corners of a triangular parking lot. Camera 1 was 156 ft from camera 2, which was 101 ft from camera 3. Cameras 1 and 3 were 130 ft apart. Which camera had to cover the greatest angle?

Answer 1

Answer:

Step-by-step explanation:

Alright, lets get started.

Please refer the diagram I have attached.

Lets use cosine law for finding the angles.

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cosA=\frac{130^2+156^2-101^2}{2*130*156}$

$cosA=0.7652$

Taking cos inverse

A = 40.08 degrees

Similarly,

$cosB=\frac{a^2+c^2-b^2}{2*a*c}$

$cosB=\frac{101^2+156^2-130^2}{2*156*101}$

$cosB=0.5597$

taking cos inverse

B = 55.97 degrees

Similarly,

$cosC=\frac{a^2 +b^2-c^2}{2ab}$

$cosC=\frac{101^2+130^2-156^2}{2*101*130}$

$cosC=0.1053$

taking cos inverse

C = 83.95 degrees

Means angle C is the largest angles,

So, camera 3 is covering the greatest angle.   :  Answer

Hope it will help :)