Question 20 5 pts

Three security cameras were mounted at the corners of a triangular parking lot. Camera 1

was 260 ft from Camera 2, which was 210 ft from Camera 3. Cameras 1 and 3 were 234 ft

apart. Which camera had to cover the greatest angle?

The camera had to cover the greatest angle is CAMERA 3 because it had the largest angle of 71.47°

Step-by-step explanation:

From the above question,

We have:

Camera 1 = Angle A

Camera 2= Angle B

Camera 3 = Angle C

A = 210ft

B = 234ft

C = 260ft

We need to find Angle A( angle of camera 1) using the cosine rule

A=(B² + C² - 2BCCosA)

210² = 234² + 260² - 2 × 234 × 260 × CosA

210² = 122356 - 121680CosA

Square both sides

210² = 122356 - 121680CosA

44100 = 122356 - 121680CosA

121680CosA = 122356 - 44100

121680CosA = 78256

Cos A = 78256/121680

Cos A = 0.6431295201

A = arc cos (0.6431295201)

A = 49.974422249°

Angle A approximately = 49.97°

Using the Sine rule to find the Angle B

A/Sine A = B/Sine B

210ft/Sine 49.97° = 234ft/Sine B

210ft × Sine B = 49.97° × 234ft

Sine B =( Sine 49.97° × 234ft)/210ft

B = arc sin (0.8532172354)

Angle B = 58.56334

Approximately = 58.56°

Angle C = 180 - (49.97 + 58.56)°

Angle C = 71.47°

Therefore, the camera had to cover the greatest angle is Camera 3 because it had the largest angle of 71.47°

the answer is c 104. i got the answers for you : )