One end of a cable is attached to the top of a flagpole and the other end is attached
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This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355