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`        In a triangle ABC, if angle C=60Then prove that 1/(a+c) + 1/(b+c) = 3/(a+b+c)Where the symbols have their usual meaning representing sides of the triangle`
2 years ago

```							Try this formula to prove reverse a^2 = b^2 + c^2 − 2bc Cos A b^2 = c^2 + a^2 − 2ca Cos B c^2 = a^2 + b^2 − 2ab Cos C u know the angle of C = 60 So c^2 = a^2 + b^2 − 2ab cosC => c^2 = a^2 + b^2 − 2ab cos 60 => c^2 = a^2 + b^2 − ab Simplify 1/(a+c) + 1/(b+c) = 3 / ( a+b+c ) => (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c) => (a+b+2c) (a+b+c) = 3(a+c) (b+c) => a^2 + b^2 + 2ab + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c^2 => a^2 + b^2 + 2ab - 3ab = 3 c^2 - 2 c^2 (cancel 3ac & 3bc) => a^2 + b^2 – ab = c^2 so both matched hence the proof
```
2 years ago
```							 1/(a+c) + 1/(b+c) = 3 / ( a+b+c ) => (a+b+2c) / (a+c) (b+c) = 3 / (a+b+c) => (a+b+2c) (a+b+c) = 3(a+c) (b+c) => a^2 + b^2 + 2ab + 2c^2 + 3ac + 3bc = 3ab + 3ac + 3bc + 3c^2 => a^2 + b^2 + 2ab - 3ab = 3 c^2 - 2 c^2 (cancel 3ac & 3bc) Now a^2+b^2-ab-c^2=0a^2+b^2-c^2=aba^2+b^2-c^2/ab=0 Divided both side by ½a^2+b^2-c^2/2ab=1/2CosC=1/2Cos C=cos 60C=60   H. P=> a^2 + b^2 – ab = c^2
```
10 months ago
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