Explanation: 
Another Explanation (5):
ABC ত্রিভুজে, দেওয়া আছে: \(\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}\)
এখন, লসাগু করে পাই,
\(\frac{b+c+a+c}{(a+c)(b+c)} = \frac{3}{a+b+c}\)
\(\implies \frac{a+b+2c}{(a+c)(b+c)} = \frac{3}{a+b+c}\)
বজ্রগুণন করে পাই,
\((a+b+2c)(a+b+c) = 3(a+c)(b+c)\)
\(\implies (a+b)^2 + 2c(a+b) + c(a+b) + 2c^2 = 3(ab + ac + bc + c^2)\)
\(\implies a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2\)
\(\implies a^2 + b^2 - ab - c^2 = 0\)
\(\implies a^2 + b^2 - c^2 = ab\)
আমরা জানি, কোসাইন সূত্রানুসারে,
\(c^2 = a^2 + b^2 - 2ab \cos{C}\)
\(\implies a^2 + b^2 - c^2 = 2ab \cos{C}\)
সুতরাং, \(ab = 2ab \cos{C}\)
\(\implies \cos{C} = \frac{1}{2}\)
\(\implies C = \cos^{-1}(\frac{1}{2})\)
\(\implies C = 60^\circ\) 🎉
অতএব, \(\angle C = 60^\circ\) 🥳
