SOLUTION: The longest side of a triangular building lot has length 190 meters and the next longer side is 180 meters (the shortest length is unknown). The angle between the longer sides meas

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Question 1178105: The longest side of a triangular building lot has length 190 meters and the next longer side is 180 meters (the shortest length is unknown). The angle between the longer sides measures 54º. Using the median to the longest side (the segment joining the midpoint of the side to the opposite vertex) the lot is divided into two lots with equal area. Find the length of the median to the nearest tenth of a meter.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
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.**

Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.
longest side of a triangular building lot has length 190 meters and the next longer side is 180 meters
(the shortest length is unknown). The angle between the longer sides measures 54º.
Using the median to the longest side (the segment joining the midpoint of the side to the opposite vertex)
the lot is divided into two lots with equal area. Find the length of the median to the nearest tenth of a meter.
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        In his solution,  @CPhill makes a lot of unnecessary calculations with no any need.
        Actually,  the problem is extremely simple and can be solved in couple of lines.
        See my solution below. 


Let A be the common vertex of the two longest sides of 190 m and 180 m.
Let B be the other endpoint of the longest, 190 m side.
Let C be the other endpoint of the 180 m side.
Let M be the midpoint of the longest, 190 m, side AB.
Then the median to the longest side is CM.


    They want you find the length of CM.


We have a triangle ACM with the sides AC = 180 m  and  AM = AB/2 = 190/2 = 95 m.
The angle CAM between these sides is 54º.


To find the length of CM, use the cosine law for triangle ACM

    CM = sqrt%28180%5E2+%2B+95%5E2+-+2%2A180%2A95%2Acos%2854%5Eo%29%29 = sqrt%28180%5E2+%2B+95%5E2+-+2%2A180%2A98%2A0.58778525229%29 = sqrt%2821322.74437%29 = 146.0230953.


ASNSWER.  The length of the median is 146.0 meters, to the nearest tenth of a meter.

Solved.

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Making unnecessary calculations does not decorate the solution,
but creates unnecessary questions and causes perplexity, instead.