Questions: Tower A Tower B 2:58 12 meters away 48.6°SW Unknown 3:12 19.2°SW 70.5°NW 3:55 10 meters away SE 29.7° NE Cell Phone Tower A and B are 20 meters apart. 1a) How far away was the phone from Tower B? 1b) How far away was the phone from Tower A? 1c) Describe the location of the phone from Tower A in terms of its angle. 2a) What store was this phone in at 2:58? 2b) What store was this phone in at 3:12? 2c) What store was this phone in at 3:55? Describe a possible route of this suspect during this time at the mall.

Solution Steps

At 2:58, the phone is 12 meters away from Tower A and the angle between Tower A and Tower B is 48.5° SW. We want to find the distance from Tower B. We know the distance between towers A and B is 20 meters. This forms a triangle. Since we have two sides and an included angle, we can use the Law of Cosines:

c² = a² + b² - 2ab * cos(C)

Where:

• c is the distance from Tower B
• a is the distance from Tower A (12 meters)
• b is the distance between towers A and B (20 meters)
• C is the angle between sides a and b (48.5°)

c² = 12² + 20² - 2 * 12 * 20 * cos(48.5°) c² = 144 + 400 - 480 * 0.6626 c² = 544 - 318.048 c² = 225.952 c ≈ 15.03 meters

At 3:12, the phone’s signal makes a 70.5° NW angle with Tower B, and a 19.2° SW angle with Tower A. We want the distance to Tower A. The distance between the towers is 20m. We need another side or angle to solve. The angle at the phone's location is 180° - 70.5° - 19.2°= 90.3°. Now we can use the Law of Sines to find the distance to Tower A:

a / sin(A) = b / sin(B)

Where:

• a is distance to Tower A (unknown)
• A is angle at Tower A (19.2°)
• b is distance between towers (20m)
• B is angle at the phone (90.3°)

a / sin(19.2°) = 20 / sin(90.3°) a = 20 * sin(19.2°) / sin(90.3°) a ≈ 20 * 0.3285 / 0.9999 a ≈ 6.57 meters

At 3:55, the phone is 10 meters away from Tower A at an angle of 29.7° NE from Tower A.

Final Answer: