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°

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.