Question 1178105
Let's solve this problem step-by-step.

**1. Visualize the Triangle**

* Let the triangle be ABC.
* Let AB = 190 meters (longest side).
* Let AC = 180 meters (next longest side).
* Let angle BAC = 54º.
* Let M be the midpoint of AB.
* We need to find the length of CM (the median).

**2. Apply the Law of Cosines to Find BC**

* BC² = AB² + AC² - 2(AB)(AC)cos(BAC)
* BC² = 190² + 180² - 2(190)(180)cos(54º)
* BC² = 36100 + 32400 - 68400cos(54º)
* BC² = 68500 - 68400(0.5878) ≈ 68500 - 40208.52
* BC² ≈ 28291.48
* BC ≈ √28291.48 ≈ 168.2 meters

**3. Find AM and MB**

* Since M is the midpoint of AB:
    * AM = MB = AB / 2 = 190 / 2 = 95 meters

**4. Apply the Law of Cosines to Triangle AMC**

* CM² = AC² + AM² - 2(AC)(AM)cos(BAC)
* CM² = 180² + 95² - 2(180)(95)cos(54º)
* CM² = 32400 + 9025 - 34200cos(54º)
* CM² = 41425 - 34200(0.5878) ≈ 41425 - 20102.76
* CM² ≈ 21322.24
* CM ≈ √21322.24 ≈ 145.9 meters

**5. Round to the Nearest Tenth**

* CM ≈ 145.9 meters

**Therefore, the length of the median to the longest side is approximately 145.9 meters.**